Trigonometric Identities and Equations: past paper questions & worked solutions

$D=8+5\sin\left(\dfrac{\pi(2)}{6}+3\right)=8+5\sin\left(\dfrac{\pi}{3}+3\right)$.

$D=4.064\ldots$, so the depth at 2 am is just over $4$ metres.

$6=8+5\sin\left(\dfrac{\pi t}{6}+3\right)$, so $\sin\left(\dfrac{\pi t}{6}+3\right)=-\dfrac25$.

After midday means $t>12$. Then $\dfrac{\pi t}{6}+3$ is greater than $2\pi+3$.

The first angle after $2\pi+3$ with sine $-\dfrac25$ is $3\pi+\sin^{-1}\left(\dfrac25\right)$.

$\dfrac{\pi t}{6}+3=3\pi+\sin^{-1}\left(\dfrac25\right)$, so $t=13.056\ldots$ hours after midnight.

$0.056\ldots$ hours is about $3.38$ minutes, so the time to the nearest minute is $13{:}03$, or $1{:}03$ pm.

Let $u=\theta+43^\circ$. The equation becomes $3\tan u=2\cos u$.

$3\dfrac{\sin u}{\cos u}=2\cos u$, so $3\sin u=2\cos^2u$.

$3\sin u=2(1-\sin^2u)$, so $2\sin^2u+3\sin u-2=0$.

$(2\sin u-1)(\sin u+2)=0$.

$\sin u=-2$ is impossible, so $\sin u=\dfrac12$.

Since $-180^\circ<\theta\leq180^\circ$, $-137^\circ<u\leq223^\circ$. Thus $u=30^\circ$ or $150^\circ$.

$\theta=u-43^\circ$, so $\theta=-13^\circ$ or $107^\circ$.

$2\tan\theta=3\cos\theta$ gives $2\dfrac{\sin\theta}{\cos\theta}=3\cos\theta$.

Since $\tan\theta$ is defined, $\cos\theta\ne0$. Hence $2\sin\theta=3\cos^2\theta$.

$2\sin\theta=3(1-\sin^2\theta)$, so $3\sin^2\theta+2\sin\theta-3=0$.

Let $\theta=2x+\dfrac\pi3$. Then $3\sin^2\theta+2\sin\theta-3=0$.

$\sin\theta=\dfrac{-2\pm\sqrt{40}}{6}=\dfrac{-1\pm\sqrt{10}}{3}$. The root $\dfrac{-1-\sqrt{10}}{3}$ is less than $-1$, so reject it.

Since $-\pi<x<\pi$, $-\dfrac{5\pi}{3}<\theta<\dfrac{7\pi}{3}$.

For $\sin\theta=\dfrac{-1+\sqrt{10}}{3}$ in this range, $\theta\approx-3.94649$, $0.804897$, $2.33670$, and $7.08808$.

Using $x=\dfrac{\theta-\pi/3}{2}$ gives $x\approx-2.49684$, $-0.121150$, $0.644749$, and $3.02044$.

To $3$ significant figures, $x=-2.50$, $-0.121$, $0.645$, and $3.02$.

$3\sin x\tan x=11+\cos x$ becomes $\dfrac{3\sin^2x}{\cos x}=11+\cos x$.

$3\sin^2x=11\cos x+\cos^2x$.

$3(1-\cos^2x)=11\cos x+\cos^2x$, so $4\cos^2x+11\cos x-3=0$.

$(4\cos x-1)(\cos x+3)=0$, so $\cos x=\dfrac14$ or $\cos x=-3$.

$\cos x=-3$ is impossible, so $\cos x=\dfrac14$.

For $0\le x<2\pi$, this gives $x=1.318\ldots$ or $x=4.965\ldots$, so $x=1.318,4.965$ radians.

Given $0<\theta<90^\circ$ and $\cos\theta=\dfrac13$, $\sin\theta=\sqrt{1-\dfrac19}=\dfrac{2\sqrt2}{3}$.

$\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{2\sqrt2/3}{1/3}=2\sqrt2$.

$2\tan\theta+3\sin\theta=0$ becomes $\dfrac{2\sin\theta}{\cos\theta}+3\sin\theta=0$.

Since $\tan\theta$ is defined, $\cos\theta\ne0$. Multiplying by $\cos\theta$ gives $2\sin\theta+3\sin\theta\cos\theta=0$.

$\sin\theta(2+3\cos\theta)=0$.

For $0<\theta\le360^\circ$, $\sin\theta=0$ gives $\theta=180^\circ$ or $360^\circ$.

$2+3\cos\theta=0$ gives $\cos\theta=-\dfrac23$, so $\theta=131.8^\circ$ or $228.2^\circ$ to one decimal place.

For $2\tan(2x+40^\circ)+3\sin(2x+40^\circ)=0$, the angle $2x+40^\circ$ must be one of the solution angles, modulo $360^\circ$.

The smallest positive value comes from $2x+40^\circ=131.8\ldots^\circ$, so $x=\dfrac{131.8\ldots-40}{2}=45.9^\circ$ to one decimal place.

$\tan\theta\sin\theta=\dfrac{\sin^2\theta}{\cos\theta}$.

$3\cos\theta\left(\tan\theta\sin\theta+3\right)=3\sin^2\theta+9\cos\theta$.

The equation becomes $3(1-\cos^2\theta)+9\cos\theta=11-5\cos\theta$.

$3-3\cos^2\theta+9\cos\theta=11-5\cos\theta$, so $3\cos^2\theta-14\cos\theta+8=0$.

For the equation in part (b), take $\theta=2x$. Then $3\cos^2(2x)-14\cos(2x)+8=0$.

Let $u=\cos(2x)$. Then $3u^2-14u+8=0$, so $(3u-2)(u-4)=0$.

$u=4$ is impossible for cosine, so $\cos(2x)=\dfrac23$.

Since $0<x<360^\circ$, we have $0<2x<720^\circ$.

In this interval, $2x=48.189\ldots^\circ,311.810\ldots^\circ,408.189\ldots^\circ,671.810\ldots^\circ$. Dividing by $2$ gives $x=24.1^\circ,155.9^\circ,204.1^\circ,335.9^\circ$ to one decimal place.

$a_1=\cos^2\left(\dfrac{\pi}{3}\right)=\left(\dfrac12\right)^2=\dfrac14$.

$a_2=\cos^2\left(\dfrac{2\pi}{3}\right)=\left(-\dfrac12\right)^2=\dfrac14$.

$a_3=\cos^2(\pi)=(-1)^2=1$.

The values of $\cos^2\left(\dfrac{n\pi}{3}\right)$ repeat every 3 terms as $\dfrac14,\dfrac14,1$.

$\sum_{n=1}^{50} n=\dfrac{50\cdot51}{2}=1275$.

There are $16$ complete groups of 3 terms in the first $48$ terms, then the first 2 terms repeat. Hence the cosine-square sum is $16\left(\dfrac14+\dfrac14+1\right)+\dfrac14+\dfrac14=24+\dfrac12=\dfrac{49}{2}$.

$\sum_{n=1}^{50}\left(n+\cos^2\left(\dfrac{n\pi}{3}\right)\right)=1275+\dfrac{49}{2}=\dfrac{2599}{2}$.

$8\tan\theta=3\cos\theta$ becomes $8\dfrac{\sin\theta}{\cos\theta}=3\cos\theta$.

Multiplying by $\cos\theta$ gives $8\sin\theta=3\cos^2\theta$.

Since $\cos^2\theta=1-\sin^2\theta$, $8\sin\theta=3(1-\sin^2\theta)$, so $3\sin^2\theta+8\sin\theta-3=0$.

For $8\tan2x=3\cos2x$, we get $3\sin^2(2x)+8\sin(2x)-3=0$.

$(3\sin2x-1)(\sin2x+3)=0$. Since $\sin2x$ cannot be $-3$, $\sin2x=\dfrac13$.

For $0\leq x\leq90^\circ$, $0\leq2x\leq180^\circ$. Thus $2x=19.471\ldots^\circ$ or $160.529\ldots^\circ$, so $x=9.74^\circ$ or $80.26^\circ$.

$3\tan\left(2x+\dfrac\pi5\right)=\sqrt3$ gives $\tan\left(2x+\dfrac\pi5\right)=\dfrac1{\sqrt3}$.

Therefore $2x+\dfrac\pi5=\dfrac\pi6+n\pi$.

$2x=-\dfrac\pi{30}+n\pi$, so $x=-\dfrac\pi{60}+\dfrac{n\pi}{2}$.

For $0\le x<\pi$, the possible values are $x=\dfrac{29\pi}{60}$ and $x=\dfrac{59\pi}{60}$.

Using $\tan\theta=\dfrac{\sin\theta}{\cos\theta}$, $5\sin\theta\tan\theta=\cos\theta+4$ becomes $\dfrac{5\sin^2\theta}{\cos\theta}=\cos\theta+4$.

Multiplying by $\cos\theta$ gives $5\sin^2\theta=\cos^2\theta+4\cos\theta$. Using $\sin^2\theta=1-\cos^2\theta$ gives $6\cos^2\theta+4\cos\theta-5=0$.

$\cos\theta=\dfrac{-2\pm\sqrt{34}}{6}$. The negative option is less than $-1$, so reject it.

$\cos\theta=\dfrac{-2+\sqrt{34}}{6}$ gives $\theta=50.320\ldots^\circ$ or $309.679\ldots^\circ$.

To one decimal place, $\theta=50.3^\circ$ or $309.7^\circ$.

$\cos\theta\left(3\tan\theta+\dfrac2{\tan\theta}\right)=3\sin\theta+\dfrac{2\cos\theta}{\tan\theta}$.

Since $\tan\theta=\dfrac{\sin\theta}{\cos\theta}$, $\dfrac{2\cos\theta}{\tan\theta}=\dfrac{2\cos^2\theta}{\sin\theta}$.

$3\sin\theta+\dfrac{2\cos^2\theta}{\sin\theta}=3\sin\theta+\dfrac{2(1-\sin^2\theta)}{\sin\theta}$.

$3\sin\theta+\dfrac2{\sin\theta}-2\sin\theta=\sin\theta+\dfrac2{\sin\theta}$, as required.

The equation becomes $\sin x+\dfrac2{\sin x}=4\sin x-5$.

Multiplying by $\sin x$ gives $\sin^2x+2=4\sin^2x-5\sin x$, so $3\sin^2x-5\sin x-2=0$.

$(3\sin x+1)(\sin x-2)=0$. Since $\sin x=2$ is impossible, $\sin x=-\dfrac13$.

For $0<x<2\pi$, $\sin x=-\dfrac13$ gives $x=\pi+\sin^{-1}\left(\dfrac13\right)$ or $x=2\pi-\sin^{-1}\left(\dfrac13\right)$.

Thus $x=3.481\ldots$ or $x=5.943\ldots$, so to $3$ significant figures $x=3.48$ or $x=5.94$.

$(3\cos\theta-\tan\theta)\cos\theta=3\cos^2\theta-\sin\theta$ because $\tan\theta\cos\theta=\sin\theta$.

The equation becomes $3\cos^2\theta-\sin\theta=2$.

Since $\cos^2\theta=1-\sin^2\theta$, $3(1-\sin^2\theta)-\sin\theta=2$.

Expanding and rearranging gives $3\sin^2\theta+\sin\theta-1=0$.

For $(3\cos2x-\tan2x)\cos2x=2$, take $\theta=2x$. Then $3\sin^2(2x)+\sin(2x)-1=0$.

Let $u=\sin(2x)$. Then $3u^2+u-1=0$, so $u=\dfrac{-1\pm\sqrt{13}}{6}$.

Since $-\dfrac{\pi}{2}\le x\le\dfrac{\pi}{2}$, we have $-\pi\le2x\le\pi$. Solving $\sin(2x)=\dfrac{-1+\sqrt{13}}{6}$ and $\sin(2x)=\dfrac{-1-\sqrt{13}}{6}$ in this interval gives $2x=-2.2665\ldots,-0.8751\ldots,0.4492\ldots,2.6924\ldots$.

Therefore $x=-1.133\ldots,-0.4375\ldots,0.2246\ldots,1.346\ldots$, giving $x=-1.13,-0.438,0.225,1.35$ radians.

Let $u=2x-15^\circ$. Then $3\sin u=\cos u$, so $\tan u=\dfrac13$.

Since $-90^\circ<x<90^\circ$, $-195^\circ<u<165^\circ$.

$u=18.434\ldots^\circ$ or $u=-161.565\ldots^\circ$, so $x=16.7^\circ$ or $x=-73.3^\circ$.

$4\sin^2\theta+8\cos\theta=3$.

Substitute $\sin^2\theta=1-\cos^2\theta$ to get $4(1-\cos^2\theta)+8\cos\theta=3$.

This gives $4\cos^2\theta-8\cos\theta-1=0$, so $\cos\theta=1\pm\dfrac{\sqrt5}{2}$.

Only $\cos\theta=1-\dfrac{\sqrt5}{2}$ lies between $-1$ and $1$.

For $0<\theta<2\pi$, $\theta=1.69$ or $4.59$ to 3 significant figures.

Starting with $\dfrac{3\sin\theta\cos\theta}{2\sin\theta-1}=5\tan\theta$, multiply by $(2\sin\theta-1)\cos\theta$.

This gives $3\sin\theta\cos^2\theta=5\sin\theta(2\sin\theta-1)$.

Substitute $\cos^2\theta=1-\sin^2\theta$: $3\sin\theta(1-\sin^2\theta)=10\sin^2\theta-5\sin\theta$.

Rearranging gives $3\sin^3\theta+10\sin^2\theta-8\sin\theta=0$.

$3s^3+10s^2-8s=0$ gives $s(3s-2)(s+4)=0$, where $s=\sin2x$.

Since $-\dfrac\pi2<2x<\dfrac\pi2$, the valid values are $\sin2x=0$ and $\sin2x=\dfrac23$.

$2x=0$ gives $x=0$. Also $2x=\sin^{-1}\left(\dfrac23\right)$, so $x=0.364863\ldots=0.365$ to 3 decimal places.

For $(2x+3)$, use $x=-\dfrac32$. Substituting gives $f\left(-\dfrac32\right)=6\left(-\dfrac32\right)^3+17\left(-\dfrac32\right)^2+4\left(-\dfrac32\right)-12=0$.

Since $f\left(-\dfrac32\right)=0$, $(2x+3)$ is a factor of $f(x)$.

$6x^3+17x^2+4x-12=(2x+3)(3x^2+4x-4)$.

$3x^2+4x-4=(3x-2)(x+2)$, so $f(x)=(2x+3)(3x-2)(x+2)$.

Let $t=\tan\theta$. Then $(2t+3)(3t-2)(t+2)=0$, so $t=-\dfrac32$, $t=\dfrac23$, or $t=-2$.

For $\dfrac\pi2<\theta<\pi$, $\tan\theta$ is negative, so use $\tan\theta=-\dfrac32$ and $\tan\theta=-2$.

This gives $\theta=\pi-\tan^{-1}\left(\dfrac32\right)=2.16$ and $\theta=\pi-\tan^{-1}(2)=2.03$, to 3 significant figures.

$4\tan\theta+5\sin\theta=0$ gives $4\dfrac{\sin\theta}{\cos\theta}+5\sin\theta=0$.

$\sin\theta\left(\dfrac4{\cos\theta}+5\right)=0$.

In $0<\theta\le360^\circ$, $\sin\theta=0$ gives $\theta=180^\circ$ or $360^\circ$.

$\dfrac4{\cos\theta}+5=0$ gives $\cos\theta=-\dfrac45$.

For $0<\theta\le360^\circ$, $\cos\theta=-\dfrac45$ gives $\theta=143.130\ldots^\circ$ or $216.869\ldots^\circ$.

The non-exact angles are $143.1^\circ$ and $216.9^\circ$ to one decimal place.

$\dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\sin x}=\dfrac5{\cos x}$. Multiplying by $\sin x\cos x$ gives $\sin^2x+\cos^2x=5\sin x$.

$\sin^2x+\cos^2x=1$, so $1=5\sin x$ and $\sin x=\dfrac15$.

For $0<x<\pi$, $x=\sin^{-1}\left(\dfrac15\right)$ or $x=\pi-\sin^{-1}\left(\dfrac15\right)$.

Thus $x=0.201357\ldots$ or $2.94023\ldots$, so $x=0.201$ or $2.94$ to $3$ significant figures.

$5\sin(3x+0.1)+2=0$ gives $\sin(3x+0.1)=-0.4$.

Since $-\dfrac{\pi}{2}<x<\pi$, multiplying by 3 and adding $0.1$ gives $-\dfrac{3\pi}{2}+0.1<3x+0.1<3\pi+0.1$.

In this interval, $3x+0.1=-2.7301\ldots,-0.4115\ldots,3.5531\ldots,5.8717\ldots$.

Subtract $0.1$ and divide by $3$: $x=-0.9433\ldots,-0.1705\ldots,1.1510\ldots,1.9239\ldots$.

To 2 decimal places, $x=-0.94,-0.17,1.15,1.92$.

$2\tan\theta\sin\theta=5+\cos\theta$. Using $\tan\theta=\dfrac{\sin\theta}{\cos\theta}$ gives $\dfrac{2\sin^2\theta}{\cos\theta}=5+\cos\theta$.

Multiplying by $\cos\theta$ gives $2\sin^2\theta=5\cos\theta+\cos^2\theta$. Substitute $\sin^2\theta=1-\cos^2\theta$.

$2(1-\cos^2\theta)=5\cos\theta+\cos^2\theta$, so $3\cos^2\theta+5\cos\theta-2=0$.

$(3\cos\theta-1)(\cos\theta+2)=0$. Since $\cos\theta=-2$ is impossible, $\cos\theta=\dfrac13$.

For $0<\theta<360^\circ$, $\cos\theta=\dfrac13$ gives $\theta=70.5^\circ$ or $289.5^\circ$ to one decimal place.

$4\tan^2(2\theta-30^\circ)+1=49$ gives $\tan^2(2\theta-30^\circ)=12$.

Let $\alpha=2\theta-30^\circ$. Since $0\leq\theta<180^\circ$, $-30^\circ\leq\alpha<330^\circ$. The solutions of $\tan\alpha=\pm\sqrt{12}$ in this range are approximately $73.9^\circ,106.1^\circ,253.9^\circ,286.1^\circ$.

$\theta=\dfrac{\alpha+30^\circ}{2}$, giving $\theta=51.9^\circ,68.1^\circ,141.9^\circ,158.1^\circ$ to one decimal place.

$2\tan x\sin x+3=0$ becomes $2\left(\dfrac{\sin x}{\cos x}\right)\sin x+3=0$.

Multiplying by $\cos x$ gives $2\sin^2 x+3\cos x=0$. Using $\sin^2x=1-\cos^2x$, $2(1-\cos^2x)+3\cos x=0$, so $2\cos^2x-3\cos x-2=0$.

$(2\cos x+1)(\cos x-2)=0$. Since $\cos x=2$ is impossible, $\cos x=-\dfrac12$. In $0\leq x<2\pi$, $x=\dfrac{2\pi}{3}$ or $x=\dfrac{4\pi}{3}$.

Let $u=2x+\dfrac\pi4$. From $\tan^2u=3$, $\tan u=\pm\sqrt3$.

Thus $u=\dfrac\pi3+n\pi$ or $u=-\dfrac\pi3+n\pi$, where $n$ is an integer.

So $x=\dfrac\pi{24}+\dfrac{n\pi}{2}$ or $x=-\dfrac{7\pi}{24}+\dfrac{n\pi}{2}$.

For $-\dfrac\pi2<x<\dfrac\pi2$, the solutions are $x=-\dfrac{11\pi}{24},-\dfrac{7\pi}{24},\dfrac\pi{24},\dfrac{5\pi}{24}$.

$(2\sin\theta-\cos\theta)^2=1$ gives $4\sin^2\theta-4\sin\theta\cos\theta+\cos^2\theta=1$.

Since $1=\sin^2\theta+\cos^2\theta$, this simplifies to $3\sin^2\theta-4\sin\theta\cos\theta=0$.

$\sin\theta(3\sin\theta-4\cos\theta)=0$.

$\sin\theta=0$ gives $\theta=180^\circ$ in the interval. Also $3\sin\theta=4\cos\theta$ gives $\tan\theta=\dfrac43$.

For $0<\theta<360^\circ$, $\tan\theta=\dfrac43$ gives $\theta=53.1^\circ$ and $233.1^\circ$. Therefore $\theta=53.1^\circ,180^\circ,233.1^\circ$.

Let $u=\theta+30^\circ$. Then $3\sin u=7\cos u$, so $\tan u=\dfrac73$.

Since $0<\theta<360^\circ$, $30^\circ<u<390^\circ$. Thus $u=66.801\ldots^\circ$ or $246.801\ldots^\circ$.

$\theta=u-30^\circ$, giving $\theta=36.8^\circ$ and $216.8^\circ$ to one decimal place.

$3\sin^3x=5\sin x-7\sin x\cos x$ gives $3\sin^3x-5\sin x+7\sin x\cos x=0$.

$\sin x(3\sin^2x-5+7\cos x)=0$.

Since $\sin^2x=1-\cos^2x$, this becomes $\sin x(-3\cos^2x+7\cos x-2)=0$, or $\sin x(3\cos^2x-7\cos x+2)=0$.

For $-\dfrac\pi2\leq x\leq\dfrac\pi2$, $\sin x=0$ gives $x=0$. Also $3\cos^2x-7\cos x+2=0$ gives $(3\cos x-1)(\cos x-2)=0$.

$\cos x=2$ is impossible, so $\cos x=\dfrac13$. The solutions are $x=-\cos^{-1}\left(\dfrac13\right)$, $0$, and $\cos^{-1}\left(\dfrac13\right)$.

$5\sin x\tan x+13=\cos x$ becomes $\dfrac{5\sin^2x}{\cos x}+13=\cos x$.

$5\sin^2x+13\cos x=\cos^2x$.

$5(1-\cos^2x)+13\cos x=\cos^2x$, so $6\cos^2x-13\cos x-5=0$.

$(2\cos x-5)(3\cos x+1)=0$, so $\cos x=\dfrac52$ or $\cos x=-\dfrac13$.

$\cos x=\dfrac52$ is impossible, so $\cos x=-\dfrac13$. For $0<x\le\pi$, $x=1.9106\ldots$, so $x=1.91$ radians to 3 significant figures.

At $t=6$, $20=10+12\sin(6k+18)^\circ$, so $\sin(6k+18)^\circ=\dfrac56$.

$\sin^{-1}\left(\dfrac56\right)=56.442\ldots^\circ$. Since $0<k<20$, $18<6k+18<138$, so $6k+18=56.442\ldots$ or $123.557\ldots$.

This gives $k=6.407\ldots$ or $k=17.592\ldots$, so $k=6.41$ or $17.59$ to 2 decimal places.

Given further that $0<k<10$, use $k=6.407\ldots$.

The maximum value of $\sin$ is $1$, so the maximum temperature is $10+12=22^\circ\text{C}$.

The maximum occurs when $kt+18=90$, so $t=\dfrac{72}{6.407\ldots}=11.2375\ldots$ hours after midnight.

$0.2375\ldots$ hours is $14.25\ldots$ minutes, so the maximum occurs at about $11{:}14$ am.

The 2nd term of $(3+2x)^6$ is $\binom61 3^5(2x)=6\cdot243\cdot2x=2916x$.

The 3rd term is $\binom62 3^4(2x)^2=15\cdot81\cdot4x^2=4860x^2$.

The 5th term is $\binom64 3^2(2x)^4=15\cdot9\cdot16x^4=2160x^4$.

For three consecutive terms in a geometric series, the middle term squared equals the product of the two outer terms.

$(4860x^2)^2=(2916x)(2160x^4)$.

For the non-zero geometric case, divide by $x^4$: $4860^2=2916\cdot2160\,x$, so $x=\dfrac{4860^2}{2916\cdot2160}=\dfrac{15}{4}$.

In the different geometric series, first term $=\sin^2\theta$, common ratio $=2\cos\theta$, and $S_\infty=\dfrac85$.

$\dfrac{\sin^2\theta}{1-2\cos\theta}=\dfrac85$.

$5\sin^2\theta=8(1-2\cos\theta)$. With $\sin^2\theta=1-\cos^2\theta$, this becomes $5(1-\cos^2\theta)=8-16\cos\theta$.

Rearranging gives $5\cos^2\theta-16\cos\theta+3=0$.

$5\cos^2\theta-16\cos\theta+3=0$ gives $\cos\theta=3$ or $\cos\theta=\dfrac15$. The possible value is $\cos\theta=\dfrac15$.

The 2nd term is first term times common ratio: $\sin^2\theta(2\cos\theta)=\left(1-\dfrac{1}{25}\right)\cdot\dfrac25=\dfrac{24}{25}\cdot\dfrac25=\dfrac{48}{125}$.

Let $u=2\theta-10^\circ$. Then $\sin u=\dfrac13$.

Since $0\leq\theta<180^\circ$, $-10^\circ\leq u<350^\circ$. In this interval, $u=19.471\ldots^\circ$ or $u=160.529\ldots^\circ$.

$\theta=\dfrac{u+10^\circ}{2}$, giving $\theta=14.7^\circ$ and $85.3^\circ$ to one decimal place.

For three consecutive terms of an arithmetic sequence, second minus first equals third minus second. Hence $\dfrac1{\tan\alpha}-\sin\alpha=2\sin\alpha-\dfrac1{\tan\alpha}$.

This gives $\dfrac2{\tan\alpha}=3\sin\alpha$. Since $\dfrac1{\tan\alpha}=\dfrac{\cos\alpha}{\sin\alpha}$, we get $\dfrac{2\cos\alpha}{\sin\alpha}=3\sin\alpha$, so $2\cos\alpha=3\sin^2\alpha$.

Using $\sin^2\alpha=1-\cos^2\alpha$, $2\cos\alpha=3(1-\cos^2\alpha)$, so $3\cos^2\alpha+2\cos\alpha-3=0$.

$\cos\alpha=\dfrac{-1+\sqrt{10}}{3}$; the other root is less than $-1$. Since $\pi<\alpha<2\pi$ and cosine is positive, $\alpha$ is in the fourth quadrant.

$\alpha=2\pi-\cos^{-1}\left(\dfrac{-1+\sqrt{10}}{3}\right)=5.517$ radians to 3 decimal places.

$\sin x\tan x=5$ becomes $\dfrac{\sin^2x}{\cos x}=5$.

Multiplying by $\cos x$ gives $\sin^2x=5\cos x$. Since $\sin^2x=1-\cos^2x$, $1-\cos^2x=5\cos x$.

$\cos^2x+5\cos x-1=0$, so $\cos x=\dfrac{-5\pm\sqrt{29}}{2}$.

$\dfrac{-5-\sqrt{29}}{2}<-1$, so the valid value is $\cos x=\dfrac{-5+\sqrt{29}}{2}$.

For $0\le x<360^\circ$, this gives $x=78.9^\circ$ or $281.1^\circ$ to one decimal place.

The point $P$ is a maximum point on the sine curve. Its $y$ coordinate is $7$, so $A+2=7$ and $A=5$.

Point $Q$ is the next minimum after $P$, so $\sin\left(2\theta-\dfrac{3\pi}{8}\right)=-1$.

For this minimum, $2\theta-\dfrac{3\pi}{8}=\dfrac{3\pi}{2}$, so $2\theta=\dfrac{15\pi}{8}$ and $\theta=\dfrac{15\pi}{16}$.

At the minimum, $y=2-5=-3$, so $Q=\left(\dfrac{15\pi}{16},-3\right)$.

At $R$, $y=0$, so $5\sin\left(2\theta-\dfrac{3\pi}{8}\right)+2=0$.

Thus $\sin\left(2\theta-\dfrac{3\pi}{8}\right)=-\dfrac25$. The crossing $R$ shown after the second maximum has $2\theta-\dfrac{3\pi}{8}=3\pi+\sin^{-1}\left(\dfrac25\right)$.

$\theta=\dfrac12\left(3\pi+\sin^{-1}\left(\dfrac25\right)+\dfrac{3\pi}{8}\right)=5.507\ldots$, so $\theta=5.51$ to 3 significant figures.