Trig Equations and R-Form: past paper questions & worked solutions

Using $\tan^2\theta=\sec^2\theta-1$ gives

The value $\sec\theta=2/3$ would give $\cos\theta=3/2$, impossible for a real angle. Thus $\cos\theta=-1/3$. In $0\lt \theta\le360^\circ$, the two angles are $\arccos(-1/3)=109.4712\ldots^\circ$ and $360^\circ-109.4712\ldots^\circ$.

The original equation requires $\sin\theta\ne0$. Using $\csc\theta=1/\sin\theta$, multiply by $\sin\theta$ to obtain $8\sin\theta\cos\theta=3$. Since $2\sin\theta\cos\theta=\sin2\theta$, this becomes $4\sin2\theta=3$.

For the smallest positive solution take the first positive sine branch: $2\theta=\arcsin(3/4)=48.59037789\ldots{}^{\circ}$. Divide by two to obtain $\theta=24.29518895\ldots{}^{\circ}$. The supplementary branch gives a larger positive angle.

Use $\cos2x=2\cos^2x-1$ and put $u=\cos x$:

The value $u=2$ is impossible since $-1\le\cos x\le1$. Thus $\cos x=-1/4$. In $0\le x\lt 360^\circ$, the solutions lie in quadrants II and III:

Let $g(x)=3x-5+5\sin(x^2)$. Using radians,

Since $g$ is continuous and changes sign, a root $\alpha$ lies between $0.7$ and $0.8$.

On the stated root interval, $0.49\lt x^2\lt 0.64\lt \pi/2$, so the principal inverse sine and positive square root give

Starting with $x_1=0.7$,

Continue $x_{n+1}=\sqrt{\arcsin(1-0.6x_n)}$ in radians, retaining the full calculator values. The iterates converge to $0.758924378\ldots$, so

Expand $R\sin(2x-\alpha)=R\cos\alpha\sin2x-R\sin\alpha\cos2x$. Comparison gives $R\cos\alpha=\sqrt3$ and $R\sin\alpha=3$. Thus $R=\sqrt{3+9}=2\sqrt3$ and $\tan\alpha=\sqrt3$. As $0\lt \alpha\lt \pi/2$, $\alpha=\pi/3$ and

The denominator is always positive, between $2\sqrt3$ and $6\sqrt3$. The fraction is smallest when the denominator is largest:

The maximum denominator occurs when $\sin(6x-\pi/3)=1$:

The smallest positive value is $x=5\pi/36$, since $n=-1$ gives a negative value.

Expand $R\cos(x-\alpha)=R\cos\alpha\cos x+R\sin\alpha\sin x$. Comparing coefficients gives $R\cos\alpha=1$ and $R\sin\alpha=2$. Thus $R^2=1^2+2^2=5$ and $\tan\alpha=2$. The specified first-quadrant phase is $\alpha=\arctan2=1.1071487\ldots$ radians.

Using the exact phase $\alpha=\arctan2$, write $g(x)=3-7\sqrt5\cos(2x-\alpha)$. (i) The negative cosine coefficient means the maximum occurs when the cosine is $-1$; the maximum is $3+7\sqrt5$. (ii) $2x-\alpha=\pi+2n\pi$, so $x=(\pi+\alpha)/2+n\pi$. The $n=-1$ value is negative; $n=0$ is the smallest positive one, $x=2.1243706857\ldots$, or 2.12 to two decimal places.

Expanding the compound angle,

so $\cos\theta+\sqrt3\sin\theta=3\sin\theta$ and $\cos\theta=(3-\sqrt3)\sin\theta$. If $\cos\theta=0$, this equation would require $\sin\theta=0$, which is impossible. We may therefore divide by $\cos\theta$:

Set $\theta=2x-10^\circ$, so $2x-70^\circ=\theta-60^\circ$ and $-10^\circ\le\theta\lt 350^\circ$.

The admissible values are $38.2619662\ldots^\circ$ and $218.2619662\ldots^\circ$. Hence $x=(\theta+10^\circ)/2$, giving $x=24.1^\circ$ or $114.1^\circ$ (1 decimal place).

Expand $R\cos(x+\alpha)=R\cos\alpha\cos x-R\sin\alpha\sin x$. Comparing coefficients gives

Hence

Thus $f(x)=25\cos(x+\alpha)$, using the unrounded $\alpha$ in subsequent calculations.

Since $-25\leq f(2x)\leq25$, the positive denominator $90-3f(2x)$ lies between $15$ and $165$. The smallest reciprocal occurs at the largest denominator:

The minimum requires $f(2x)=-25$, so

Since $0\lt \alpha\lt \pi/2$, the first positive solution uses $n=0$:

Since $2e^{3\sin x}\gt 0$, zeros occur when $\cos x=0$. The labelled point R is the right-hand crossing, at $x=3\pi/2$.

For $f(x)=2e^{3\sin x}\cos x$, the product and chain rules give

At a turning point the derivative is zero. The exponential factor is nonzero, so $3\cos^2x-\sin x=0$. Substitute $\cos^2x=1-\sin^2x$ and rearrange: $3-3\sin^2x-\sin x=0$.

Solving $3s^2+s-3=0$ for $s=\sin x$ gives

The negative root is less than $-1$, so it is impossible for a real sine. Let $\beta=\arcsin((\sqrt{37}-1)/6)$ in the first quadrant. The labelled minimum Q lies in the second quadrant, hence $x_Q=\pi-\beta=2.131037293\ldots$.

The zero-product rule gives $x-2=0$ or $\sqrt3\sec x+2=0$. The first gives $x=2$, which lies in $(0,\pi)$ and has $\cos x\ne0$. The second gives $\cos x=-\sqrt3/2$; the only angle in the interval is $5\pi/6$.

Use $\cos2\theta=1-2\sin^2\theta$ and set $s=\sin\theta$:

The negative root is less than $-1$, so reject it. The positive root is $0.2595730874\ldots$. Its principal angle is $15.04473218\ldots^\circ$; the other solution is $180^\circ-15.04473218\ldots^\circ$. Both are in the given interval. To three significant figures:

Expand the compound angle using exact values:

Thus $\sqrt3\sin\theta=6\cos\theta$. If $\cos\theta=0$, this equation is impossible, so divide by $\sqrt3\cos\theta$: $\tan\theta=6/\sqrt3=2\sqrt3$.

Put $\theta=x+20^{\circ}$, so $x-10^{\circ}=\theta-30^{\circ}$. Part (a) gives $\tan(x+20^{\circ})=2\sqrt3$. Since $20^{\circ}\le\theta\le380^{\circ}$, the two permitted values are $\theta=73.89788625\ldots{}^{\circ}$ and $253.89788625\ldots{}^{\circ}$. Subtract $20^{\circ}$ from both and round last.

Expand and use $\cos^22x=(1+\cos4x)/2$:

The zeros satisfy $\cos2x=-1/2$. The first two positive values are $x=\pi/3$ and $x=2\pi/3$, so $a=2\pi/3$. The squared curve is nonnegative, so integrate directly for the total shaded area:

Use $\sin(4\pi/3)=-\sqrt3/2$ and $\sin(8\pi/3)=\sqrt3/2$ to obtain $\mathcal A=2\pi-\sqrt3+\sqrt3/4=2\pi-3\sqrt3/4$.

Using the product and chain rules,

At a stationary point, divide by $x/2$, since $x\gt 0$, to obtain $4\cos(x/2)=x\sin(x/2)$. The endpoint $x=\pi$ does not satisfy this equation, so at P we have $0\lt x\lt \pi$ and $\cos(x/2)\gt 0$. Hence $\tan(x/2)=4/x$. The principal arctangent is valid because $0\lt x/2\lt \pi/2$:

Use radian mode and retain full calculator precision between iterations:

Thus the requested second and sixth terms, not the limiting value, are as below.

Expanding the required form gives

Comparing coefficients with \(5\cos\theta-4\sin\theta\),

Squaring and adding, using \(R\gt 0\), gives

Also, since \(0\lt \alpha\lt \frac\pi2\),

Thus the exact form is

A stretch parallel to the \(y\)-axis with scale factor \(\boxed{\sqrt{41}}\), taking \(y=\cos\theta\) to \(y=\sqrt{41}\cos\theta\).

A translation by the vector

This moves the curve \(\alpha\) units to the left, giving \(y=\sqrt{41}\cos(\theta+\alpha)\).

Since \(\theta\) can be any real number,

Hence \(4\leq4+(f(\theta))^2\leq45\). Taking reciprocals of these positive bounds and multiplying by \(90\) gives

Both endpoints are attained: the minimum occurs when \(f(\theta)=\pm\sqrt{41}\), and the maximum when \(f(\theta)=0\).

Put $s=\sin x$ and $c=\cos x$. Using the addition and double-angle formulae,

Where the stated expressions are defined, $c\ne0$ and $\cos3x\ne0$. Dividing numerator and denominator by $c^3$ gives

By part (a), the equation is $\tan3\theta=2\sec^2 3\theta-8$. Set $u=\tan3\theta$ and use $\sec^2 3\theta=1+u^2$:

Thus $\tan3\theta=2$ or $\tan3\theta=-3/2$. The transformed interval is $0\lt 3\theta\lt 3\pi/2$. Its three allowed angles are

Consequently

None lies at an excluded pole of either side.

Expand $\sin(2x+x)$ using the addition formula:

Now use $\cos^2x=1-\sin^2x$:

Substitute the identity and $\sin2\theta=2\sin\theta\cos\theta$:

Factor without dividing by $\sin\theta$:

If $\sin\theta=0$, the permitted solutions are $180^\circ$ and $360^\circ$ (not $0^\circ$). Otherwise $\cos\theta=(5\pm\sqrt{41})/8$. The plus value exceeds 1 and is rejected. The minus value gives $\theta=100.1013\ldots^\circ$ or $259.8986\ldots^\circ$.

Expand both compound angles:

Thus $(2-\sqrt3)\sin x=-\cos x$. Here $\cos x\ne0$: otherwise this equation would require $\sin x=0$ as well, which is impossible. Dividing by $(2-\sqrt3)\cos x$ gives

Set $x=2\theta-45^\circ$. Then $x+45^\circ=2\theta$ and $x-60^\circ=2\theta-105^\circ$, so part (a) gives $\tan(2\theta-45^\circ)=-2-\sqrt3$. Since $\tan75^\circ=2+\sqrt3$,

In $0\le\theta\lt 180^\circ$, only $k=1,2$ are allowed. Substitution in the original equation confirms both values.

The original equation is defined only when $\cos\theta+\sin\theta\ne0$ and $\cos2\theta\ne0$. Multiplying by $\cos\theta+\sin\theta$ gives

Using $2\sin\theta\cos\theta=\sin2\theta$, multiply through by 2: $3\sin2\theta=4\cos2\theta+2$.

Write $3\sin2x-4\cos2x=5\sin(2x-\phi)$, where $\cos\phi=3/5$, $\sin\phi=4/5$ and $\phi=\arctan(4/3)$. Then $\sin(2x-\phi)=2/5$. Because $\pi\lt x\lt 3\pi/2$, $2\pi-\phi\lt 2x-\phi\lt 3\pi-\phi$, approximately $(5.356,8.497)$. With $\gamma=\arcsin(2/5)$, the only angle in this interval is $2\pi+\gamma$; $3\pi-\gamma\approx9.013$ is outside it. Thus

At this value $\cos x+\sin x\ne0$ and $\cos2x\ne0$; direct substitution satisfies the original equation.

Starting from the left side and using double-angle identities,

These steps hold wherever the original expression is defined, so $\sin2\theta\ne0$; no claim is made at the excluded angles.

By part (a), the equation becomes

The branch $\sec x=4$ gives $\cos x=1/4$ and hence $x=75.5225\ldots^\circ$ or $284.4775\ldots^\circ$. The other branch gives $\cos x=-1$, hence $x=180^\circ$ in the stated interval. Reject $180^\circ$: it is a multiple of $90^\circ$ and the original $\csc2x$ is undefined there. The two remaining angles satisfy the exclusions.

On the domain of the original equation, multiply by $\cos\theta$:

Divide by the nonzero $\sin2\theta$ and use $\cot^2u=\csc^2u-1$:

Collect the terms to obtain the displayed equation.

Put $u=\csc2\theta$. Then $(3u+2)(u-5)=0$, so $u=5$ or $-2/3$. The second would give $\sin2\theta=-3/2$, which is impossible. For $0\lt \theta\lt \pi/2$, we have $0\lt 2\theta\lt \pi$. Thus

Hence $\theta=0.1006789604\ldots$ or $1.4701173664\ldots$ radians. Both satisfy the original denominator restrictions.

Using $\sec^2x=1+\tan^2x$, the equation becomes $2\tan^2x-3\tan x=0$, or $\tan x(2\tan x-3)=0$. If $\tan x=0$, the given interval $0\lt x\le\pi$ contains only $x=\pi$. If $\tan x=3/2$, the sole root in this interval is $x=\arctan(3/2)=0.982793723\ldots$ radians. The excluded pole $x=\pi/2$ supplies no solution.

Use angle addition to expand:

Where $\sin\theta\cos\theta\ne0$, divide by the stated denominators:

Expand $R\sin(x-\alpha)=R\cos\alpha\sin x-R\sin\alpha\cos x$. Therefore $R\cos\alpha=12$, $R\sin\alpha=5$. Thus $R=\sqrt{12^2+5^2}=13$ and $\alpha=\arctan(5/12)=0.3947911197\ldots$ radians, in the required first quadrant.

Write $g(\theta)=10+13\sin(2\theta-\pi/6-\alpha)$. (i) The minimum is $10-13=-3$. (ii) For sine to equal $-1$, $2\theta-\pi/6-\alpha=-\pi/2+2\pi n$. Hence $\theta=-\pi/6+\alpha/2+\pi n$. For $n=0$ this is negative, so the smallest permitted value uses $n=1$:

Use the unrounded phase angle for the calculation.

As $\beta$ varies over the reals, $12\sin\beta-5\cos\beta$ covers $[-13,13]$. Its square therefore covers $[0,169]$, not $[169,169]$. Subtracting from 10 reverses the bounds, giving $[-159,10]$; both endpoints are attained.

By the product and chain rules, $f'(x)=\cos(x/3)-(x/3)\sin(x/3)$.

Set $f'(x)=0$: $\cos(x/3)=(x/3)\sin(x/3)$. At a solution, cosine cannot vanish (that would leave a nonzero right side), so $\tan(x/3)=3/x$. On the principal positive branch $0\lt x/3\lt \pi/2$, this gives

This is the branch used by the supplied iteration for the root near 2.581. The original domain $x\gt 0$ also admits later tangent branches: in general $x=3\arctan(3/x)+3n\pi$. The principal-branch rearrangement must not be interpreted as listing every positive stationary point.

Using radians and $x_1=2.5$, the successive values are

Keep full precision between steps and round the requested terms only.

Define $F(x)=\cos(x/3)-(x/3)\sin(x/3)$, which is continuous near 2.581. The rounding interval endpoints give

By continuity a root lies strictly between these endpoints; every number in this open interval rounds to 2.581 to three decimal places.

Let $t=\tan x$. For the original equation to be defined, $\sin x\ne0$, $\cos x\ne0$ and $\cos 2x\ne0$; hence $t\ne0$ and $t^2\ne1$. Using the double-angle and reciprocal identities,

Substituting into the equation gives

Multiply by the nonzero quantity $t(1-t^2)$ and expand:

Therefore

Factor the quadratic in $t^2$:

Since $t$ is real, $t^2\ge0$, so discard $t^2=-1/5$. Thus

In degrees, let

Tangent is positive in quadrants I and III and negative in quadrants II and IV. The four solutions in $0\le x\lt 360^\circ$ are therefore

These have finite, nonzero tangent with $t^2=5\ne1$, so every transformation in part (a) is reversible and all four satisfy the original equation. To one decimal place,

For angles where the original expression is defined,

Divide the given equation by $4\operatorname{cosec}4\theta$ and use (a):

Use $\cos^2\theta=(1+\cos2\theta)/2$ and multiply by $2\cos2\theta$:

Therefore

The divisions are valid on the domain of the original equation.

Set $u=\cos2x$:

Reject $u=3$ because cosine lies between $-1$ and $1$. Thus $\cos2x=-2/5$. Since $0\lt 2x\lt 360^\circ$,

Hence $\boxed{x=56.8^\circ\text{ or }123.2^\circ}$ to one decimal place. Neither is an excluded multiple of $45^\circ$; these are all the solutions in the interval.

Use $2\sin x\cos x=\sin2x$ and $2\sin^2x=1-\cos2x$:

Expanding $R\sin(2x+\alpha)$ and comparing coefficients gives

Therefore

Thus $h(x)=5+\frac{3\sqrt5}{2}\sin(2x+\alpha)$ with the exact angle $\alpha=\arctan(2)$.

The given interval includes two complete periods. The range is

At a minimum, $\sin(2x+\alpha)=-1$, so

On $0\le x\le2\pi$, $n=0$ and $n=1$ give the two minima. The larger coordinate is

Use $2\sin^2\theta=1-\cos2\theta$, $\cot2\theta=\cos2\theta/\sin2\theta$, and $2\sin\theta\cos\theta=\sin2\theta$:

This is in terms of $\sin2\theta$ and $\cos2\theta$ only, on the original domain $\sin2\theta\ne0$.

Substitute the preceding expression and multiply by $\sin2\theta$:

Cancel $3\cos2\theta$ and use $\cos^22\theta=1-\sin^22\theta$:

Hence $5\sin^22\theta+14\sin2\theta-3=0$.

Let $s=\sin2x$. Then $(5s-1)(s+3)=0$, so $s=1/5$ or $s=-3$. The latter is impossible. As $0\lt 2x\lt 180^\circ$,

Therefore $x=5.8^\circ$ or $84.2^\circ$ to one decimal place. Both have nonzero $\sin2x$, so the original cotangent is defined.

Apply the product and chain rules:

Since $e^{-x^2}\gt 0$, a stationary point satisfies

Neither endpoint $x=0,\pi/3$ is stationary. In the interior $\sin3x\gt 0$ and $x\gt 0$, so $\cos3x\gt 0$ at a stationary point and $0\lt 3x\lt \pi/2$. Therefore division and the principal inverse tangent are valid:

Work in radians and retain calculator precision between iterations:

Thus (i) $\boxed{x_2=0.4367}$ and (ii) $\boxed{x_4=0.4307}$.

Let $h(x)=3\cos3x-2x\sin3x$, which is continuous. At the rounding boundaries,

The sign change places a root strictly between $0.4295$ and $0.4305$. On $0\lt x\lt \pi/6$, $h'(x)=-11\sin3x-6x\cos3x\lt 0$, so the stationary root is unique; for $\pi/6\leq x\leq\pi/3$, $h(x)\lt 0$. Hence the $x$ coordinate of $P$ is $\boxed{0.430\text{ to 3 decimal places}}$.

Use $\csc\theta=1/\sin\theta$. Since $0\lt \theta\lt \pi$, $\sin\theta\ne0$ and

Thus $\sin2\theta=3/4$. Since $0\lt 2\theta\lt 2\pi$, the two possibilities are

They give $\theta=0.4240310395\ldots$ and $1.1467652873\ldots$ radians.

Apply the tangent subtraction identity:

Set $u=2x-70^\circ$. The interval becomes $-70^\circ\lt u\lt 290^\circ$. With $\arctan(-3/8)=-20.5560452196\ldots^\circ$, the permitted values are $u=-20.5560452196\ldots^\circ$ and $159.4439547804\ldots^\circ$. Adding $70^\circ$ and halving gives $x=24.7219773902\ldots^\circ$ and $114.7219773902\ldots^\circ$. Both keep the original tangents and denominator defined.

Differentiate $x=3\cos2y$ with respect to $y$ using the chain rule:

For $0\lt y\lt \pi/2$, $\sin2y\gt 0$ and

Since $\cos2y=x/3$ and the sine is positive on this interval,

Consequently $dy/dx=-1/(2\sqrt{9-x^2})$, valid for $-3\lt x\lt 3$. The endpoint tangents are vertical; there is no finite derivative there.

Set the gradient equal to $-1/4$:

Because $a\gt 0$, take $a=\sqrt5$. Substitute into $a=3\cos2b$:

Here $0\lt 2b\lt \pi/2$ since both coordinates are positive, so the principal arccos value is the required one.

Use $\sin2x=2\sin x\cos x$ and $2\cos^2x=1+\cos2x$:

Expanding $R\sin(2x+\alpha)$ and matching coefficients gives

Squaring and adding, $R^2=20$, so $R=2\sqrt5$ since $R\gt 0$. Also $\tan\alpha=1/2$, and $0\lt \alpha\lt \pi/2$ selects $\alpha=\arctan(1/2)=0.463647609\ldots$ radians. Thus $f(x)=2\sqrt5\sin(2x+\arctan(1/2))-1$. With the phase displayed to 3 significant figures, $f(x)\approx2\sqrt5\sin(2x+0.464)-1$.

(i) The sine term has maximum 1, hence the maximum of $f$ is $2\sqrt5-1$. (ii) Maxima occur when

The value for $k=0$ is $0.553574\ldots\gt 0$, while $k=-1$ is negative. The second positive value therefore has $k=1$:

At $x=1/4$, $\sin^2 4y=1/4$. Since $0\le4y\le\pi/2$,

Only this branch lies in the given range of $y$.

Differentiate $x=(\sin4y)^2$ with respect to $y$:

On the interior of the given interval, $\sin4y=\sqrt{x}$ and $\cos4y=\sqrt{1-x}$ are positive. Taking the reciprocal,

Complete the square:

Thus $dy/dx=(q+r(x+s)^2)^{-1/2}$ has the constants below, with $0\lt x\lt 1$ for a finite derivative; the endpoints have vertical tangents.

The derivative is positive. Its minimum occurs when the square root in the denominator is largest. Since $(x-1/2)^2\ge0$, the largest radicand is 16, attained at $x=1/2$. Hence the smallest gradient is $1/\sqrt{16}$.

For $x\ne n\pi/2$, both sine and cosine are nonzero. Then

By part (a), the equation is $\cot^2\theta-\tan^2\theta=2\tan^2\theta$, so $\cot^2\theta=3\tan^2\theta$. The original expression excludes $\theta=0$ in the stated interval, so $\tan\theta\ne0$ and multiplication by $\tan^2\theta$ gives

Tangent is one-to-one on $(-\pi/2,\pi/2)$; the two angles are $\pm\arctan(3^{-1/4})=\pm0.6497662865\ldots$ radians. Both are valid and nonzero. To two decimal places:

Expand $R\sin(x-\alpha)=R\cos\alpha\sin x-R\sin\alpha\cos x$. Thus $R\cos\alpha=8$, $R\sin\alpha=15$, so $R=\sqrt{64+225}=17$ and $\alpha=\arctan(15/8)=1.0808390005\ldots$ radians, in the required quadrant.

Using the exact phase from (a),

The denominator lies between 7 and 75, so it is always positive. (i) The smallest fraction occurs at its largest denominator: $f_{\min}=15/75=1/5$. (ii) This occurs when $x-\alpha=\pi/2+2n\pi$. The smallest positive value is $x=\alpha+\pi/2=2.6516353273\ldots$, or 2.65 to three significant figures.

The transformation $2f(x)-5$ preserves the locations of minima and maps their y coordinate to $2(1/5)-5=-23/5$.

Reflection in the x-axis changes minima of f into maxima of $-f$. The argument $2x$ halves their x coordinates, so the first positive maximum is at

Write $s=\sin\theta$. The compound-angle formula gives

Hence

The given nonzero denominator ensures that $1-s\ne0$ and the other factor is nonzero. Cancel the common factor to obtain the required identity.

Use (a) and $\csc\theta=1/\sin\theta$. With $s=\sin\theta\ne0$,

Thus $s=(-5\pm\sqrt{97})/18$, approximately $0.269380989$ and $-0.824936545$. Both lie in $[-1,1]$, are nonzero, and satisfy the original nonzero-denominator condition. The positive value gives $15.6274\ldots^\circ$ and $164.3726\ldots^\circ$; the negative value gives $235.5821\ldots^\circ$ and $304.4179\ldots^\circ$. All four angles lie in $0\lt \theta\le360^\circ$.

Differentiate by the quotient rule:

Inside the stated interval, $1+\sin x\gt 0$. At the interior minimum M, $y'=0$, so its numerator is zero, giving the required equation.

Write $2\sin x+\cos x=\sqrt5\sin(x+\beta)$, where $\cos\beta=2/\sqrt5$, $\sin\beta=1/\sqrt5$ and $\beta=\arctan(1/2)$. The equation becomes $\sin(x+\beta)=-2/\sqrt5$. Let $\gamma=\arcsin(2/\sqrt5)$, so $\gamma+\beta=\pi/2$. The solutions in the closed boundary range include $x=-\pi/2$, $3\pi/2$, and

Both boundary values are outside the original open domain (and make its denominator zero). The sole interior stationary point is M; the derivative changes from negative to positive there.

Use $1-\cos2x=2\sin^2x$ and $\sin2x=2\sin x\cos x$:

The source excludes multiples of $90^\circ$, so the cancelled sine and denominator cosine are nonzero.

Part (a) reduces the equation to $(9/2)\tan\theta=2\sec^2\theta$. Set $u=\tan\theta$ and use $\sec^2\theta=1+\tan^2\theta$:

Both roots are positive. On $0\lt \theta\lt 90^\circ$, tangent is one-to-one, so each gives exactly one angle: $31.36697777\ldots{}^\circ$ and $58.63302223\ldots{}^\circ$.

Apply the double-angle identities to the left side:

All divisions and cancellations are valid under the given exclusion $x\ne n\pi/2$.

Using (a) with $x=2\theta$ gives $7+\csc2\theta=3\cot^22\theta$. Set $u=\csc2\theta$ and use $\cot^22\theta=u^2-1$: $3u^2-u-10=(3u+5)(u-2)=0$. Thus $\sin2\theta=1/2$ or $\sin2\theta=-3/5$. As $-\pi\lt 2\theta\lt \pi$, the first gives $2\theta=\pi/6,5\pi/6$. Write $\gamma=\arcsin(3/5)$; the second gives $2\theta=-\gamma,-\pi+\gamma$. Therefore the four values are $\pi/12,5\pi/12,-0.3217505544\ldots,-1.249045772\ldots$. None makes an original sine or cosine denominator zero.

Expand $R\cos(x-\alpha)=R\cos\alpha\cos x+R\sin\alpha\sin x$. Matching coefficients gives $R\cos\alpha=1$, $R\sin\alpha=4$. Thus $R=\sqrt{1^2+4^2}=\sqrt{17}$ and $\tan\alpha=4$. Since $0\lt \alpha\lt \pi/2$, $\alpha=\arctan4=1.32581766\ldots$ radians.

The model becomes $H=24/[3+\sqrt{17}\cos(t/2-\alpha)]$. For $0\le t\le6.5$, the phase lies between $-\alpha$ and $3.25-\alpha$; throughout this interval the denominator remains positive. Its maximum $3+\sqrt{17}$ is attained at $t=2\alpha=2.651635\ldots$, inside the time interval. The height is then smallest:

To the nearest centimetre this is $3.37\ \mathrm m$ (337 cm).

For $H=10$, $3+\sqrt{17}\cos(t/2-\alpha)=24/10$, so

Let $\gamma=\arccos[-3/(5\sqrt{17})]$. In the permitted phase interval $[-\alpha,3.25-\alpha]$, only $+\gamma$ occurs: $-\gamma\lt -\alpha$, and shifts by $2\pi$ are outside it. Using the unrounded $\alpha=\arctan4$,

By the product and chain rules,

For $-\pi/4\lt x\lt \pi/4$, $e^{3x}\sec2x$ is positive and nonzero. Hence $g'(x)=0$ precisely when $3+2\tan2x=0$, or $\tan2x=-3/2$. Since $-\pi/2\lt 2x\lt \pi/2$, tangent is one-to-one there:

Differentiate $x=\ln(\sin y)$ with respect to y:

The original equation gives $\sin y=e^x$. Since $0\lt y\lt \pi/2$, cosine is positive, so $\cos y=\sqrt{1-\sin^2y}=\sqrt{1-e^{2x}}$. Therefore

Here $x\lt 0$ from the original interval, so the square root is real and nonzero.

Using a common denominator,

The stated exclusion $x\ne n\pi/2$ ensures both denominators are nonzero.

By part (a), the squared bracket is $\csc^2\theta$. Using $\csc^2\theta=1+\cot^2\theta$ and setting $u=\cot\theta$ gives

On $0\lt \theta\lt \pi/2$, both positive cotangents are permitted: $u=1$ gives $\theta=\pi/4$, while $u=5$ gives $\tan\theta=1/5$, $\theta=0.1973955598\ldots$ radians.

The bracket in part (a) equals $\csc x$. Therefore

On the $y$ axis, $x=0$, so $\sin(3y+\pi/4)=0$. Since $\pi/12\lt y\lt 3\pi/4$, we have $\pi/2\lt 3y+\pi/4\lt 5\pi/2$. Therefore the possible angles are $\pi$ and $2\pi$, giving

Differentiate $x$ with respect to $y$, then take the reciprocal:

Using $\sin(3y+\pi/4)=3x/2$,

Thus $\boxed{p=4,\ q=9}$. The derivative formula applies where the tangent is not vertical.

At $x=0$, part (b) gives $|dy/dx|=1/2$. The sketch shows a negative tangent gradient at $A$ and a positive one at $B$. The normal at $A$ therefore has gradient $2$, while the tangent at $B$ has gradient $1/2$:

At their intersection,

Let $f(x)=2\cos3x-3x+4$, using radians. Then

The function is continuous on $[0.8,0.9]$, so this change of sign shows that there is a root between the endpoints. Moreover, $f'(x)=-6\sin3x-3\lt 0$ on this interval, so this root is unique there. Thus the crossing shown at $P$ has $\boxed{0.8\lt \alpha\lt 0.9}$.

Using the inverse cosine in radians,

Hence $\boxed{x_2=0.8327}$ to 4 decimal places.

Continue the iteration, retaining the full calculator values at each step:

Therefore $\boxed{x_5=0.8110}$ to 4 decimal places.

Differentiate the curve equation:

At a stationary point,

For $x\gt 0$, the stationary points therefore belong to the two families

In the first family, $\cos3x=-\frac{\sqrt3}{2}$, giving $y”=9\sqrt3\gt 0$, so these are local minima. In the second family, $\cos3x=\frac{\sqrt3}{2}$ and $y”=-9\sqrt3\lt 0$, so these are local maxima and are excluded. Thus all positive minimum coordinates are

The two smallest are obtained with $n=0$ and $n=1$: