The Scenario
Not every trigonometric equation is a one-step solve. Consider
$$\cos 2\theta+3\sin\theta=2, \qquad 0\leqslant\theta\leqslant 2\pi.$$
It mixes $\cos 2\theta$ with $\sin\theta$ โ different angles, different functions. The key is to choose the right identity first so that everything is expressed in one function of one angle.
Review ยท Activating Prior Knowledge
Quick check โ pick the correct statement in each question before we solve today's trig equation.
$\cos 2\theta=1-2\sin^2\theta$ expresses $\cos 2\theta$ using only $\sin\theta$. (The equivalent cosine-only form is $2\cos^2\theta-1$.) Substituting this into $\cos 2\theta+3\sin\theta=2$ turns everything into a quadratic in $\sin\theta$.
Treat the trig function as the "unknown": solve the quadratic for $\sin\theta$ (or $\cos\theta$, $\tan\theta$) first, then solve for the angle within the required interval. Never divide by the trig function unless you have checked it cannot be zero.
Because the equation already contains $3\sin\theta$, the double-angle form $\cos 2\theta=1-2\sin^2\theta$ rewrites it as $2\sin^2\theta-3\sin\theta+1=0$ โ a quadratic in one function of one angle. That is the goal of the whole toolkit.
The Question
For $\cos 2\theta+3\sin\theta=2$, which single identity converts the whole equation into a quadratic in $\sin\theta$ โ and what is the resulting quadratic? Then outline the remaining solve steps.
๐ก The Bridge (click to reveal)
Substituting $\cos 2\theta=1-2\sin^2\theta$ gives $1-2\sin^2\theta+3\sin\theta=2$, i.e. $2\sin^2\theta-3\sin\theta+1=0$. Factorising: $(2\sin\theta-1)(\sin\theta-1)=0$, so $\sin\theta=\tfrac12$ or $\sin\theta=1$, then solve over $[0,2\pi]$. Today we combine all our tools on such multi-step equations.
Every trigonometric equation you will meet at A-level can be solved by one (or a chain) of three strategies. The skill is recognising which strategy an equation needs, and chaining them when one alone is not enough.
If the equation mixes $\cos 2\theta$ with $\sin\theta$, or mixes $\sin^2\theta$ with $\cos\theta$, use a double-angle identity to rewrite everything in terms of one function.
$$\cos 2\theta = 1 - 2\sin^2\theta \;=\; 2\cos^2\theta - 1 \;=\; \cos^2\theta - \sin^2\theta$$
After substitution, the equation becomes a polynomial in $\sin\theta$ (or $\cos\theta$) that you can factorise.
If the equation is (or reduces to) $A u^2 + B u + C = 0$ where $u = \sin\theta$, $\cos\theta$, or $\tan\theta$, factorise it as $(pu + q)(ru + s) = 0$ and solve each linear factor separately.
Example: $\tan^2\theta - 3\tan\theta - 4 = 0 \;\Rightarrow\; (\tan\theta - 4)(\tan\theta + 1) = 0 \;\Rightarrow\; \tan\theta = 4 \text{ or } \tan\theta = -1.$
When the equation has a linear combination of $\cos\theta$ and $\sin\theta$ (not squared), convert to a single sinusoid:
$$a\cos\theta + b\sin\theta = R\cos(\theta - \alpha), \qquad R = \sqrt{a^2 + b^2}, \quad \tan\alpha = \dfrac{b}{a}$$
Then solve $R\cos(\theta - \alpha) = c$, i.e. $\cos(\theta - \alpha) = \dfrac{c}{R}$, taking care with the quadrant and the interval.
Chaining strategies: A single equation may need two strategies in sequence. The most common chain is Strategy 1 then Strategy 2: use an identity to convert $\cos 2\theta$ into $\sin^2\theta$, then factorise the resulting quadratic in $\sin\theta$.
Work through six steps in order. At each step, first discuss and predict with your partner, then tap Next to reveal the worked transformation and check your idea.
Look at $\cos 2\theta + 3\sin\theta = 2$. It mixes a doubled angle, $\cos 2\theta$, with a plain $\sin\theta$. Without solving: which of your three strategies is the first move โ and is one strategy enough, or do you need to chain two?
The goal is one function of one angle. Of the three forms โ $\cos 2\theta = 1-2\sin^2\theta$, $\cos 2\theta = 2\cos^2\theta-1$, or $\cos 2\theta = \cos^2\theta-\sin^2\theta$ โ which produces only $\sin\theta$? (Predict, then reveal.)
Replace $\cos 2\theta$ and rearrange so the equation reads $0 = \dots$ (Predict the new shape, then reveal.)
With $u=\sin\theta$ it reads $2u^2-3u+1=0$. Which strategy applies now? Factorise it. (Predict, then reveal.)
For $0\leq\theta\leq2\pi$, solve $\sin\theta=\tfrac12$ and $\sin\theta=1$. (Predict, then reveal.)
List the strategies you used, in the order you used them. Was any single strategy enough on its own?
Five equations appear one at a time. For each, pick the strategy you would use FIRST, tap Check โ right or wrong, the why is revealed โ then Next brings the following equation.
- Equation mixes $\sin 2\theta$ / $\cos 2\theta$ with $\theta$? โ double-angle identity (A), then factorise (B).
- Quadratic in one trig function? โ factorise (B).
- A linear combination $a\cos\theta+b\sin\theta$ (not squared)? โ R-form (C).
- A single sinusoid already isolated? โ solve it directly (D).
Discussion: Compare your tree with a neighbour. Which of your classifications turned out to need a chain of strategies rather than a single one?
Solve the following equations.
(i) $\sin( \theta + 40^\circ) = 0.7 \qquad 0^\circ \leqslant \theta \leqslant 360^\circ$
(ii) $3\cos^2 \theta + 5\sin \theta - 1 = 0 \qquad 0^\circ \leqslant \theta \leqslant 360^\circ$
๐ Reveal Solution
Let $u = \theta + 40^{\circ}$. When $0^{\circ} \leq \theta \leq 360^{\circ}$, we have $40^{\circ} \leq u \leq 400^{\circ}$.
$\sin u = 0.7$. Principal value: $u_0 = \arcsin(0.7) \approx 44.42^{\circ}$.
Second solution in one period: $u = 180^{\circ} - 44.42^{\circ} = 135.58^{\circ}$.
Add $360^{\circ}$ to find further solutions within $[40^{\circ}, 400^{\circ}]$: $u = 44.42^{\circ} + 360^{\circ} = 404.42^{\circ}$ (out of range); $u = 135.58^{\circ}$ (in range); $u = 135.58^{\circ} + 360^{\circ} = 495.58^{\circ}$ (out of range).
Convert back: $\theta = u - 40^{\circ}$.
$\theta = 44.42^{\circ} - 40^{\circ} = 4.42^{\circ}$, and $\theta = 135.58^{\circ} - 40^{\circ} = 95.58^{\circ}$.
$\boxed{\theta \approx 4.42^{\circ},\; 95.58^{\circ}}$ (2 d.p.)
Use $\cos^2\theta = 1 - \sin^2\theta$:
$3(1 - \sin^2\theta) + 5\sin\theta - 1 = 0$
$3 - 3\sin^2\theta + 5\sin\theta - 1 = 0$
$-3\sin^2\theta + 5\sin\theta + 2 = 0$
$3\sin^2\theta - 5\sin\theta - 2 = 0$
Factorise: $(3\sin\theta + 1)(\sin\theta - 2) = 0$.
$\sin\theta = -\dfrac{1}{3}$ or $\sin\theta = 2$ (reject, since $|\sin\theta| \leq 1$).
For $\sin\theta = -\dfrac{1}{3}$: principal value $\theta_0 = \arcsin(-\tfrac{1}{3}) \approx -19.47^{\circ}$. In $[0^{\circ}, 360^{\circ}]$, sine is negative in Q3 and Q4:
$\theta = 180^{\circ} + 19.47^{\circ} = 199.47^{\circ}$ (Q3), and $\theta = 360^{\circ} - 19.47^{\circ} = 340.53^{\circ}$ (Q4).
$\boxed{\theta \approx 199.47^{\circ},\; 340.53^{\circ}}$ (2 d.p.)
Solve $\cos 2\theta + 3 \sin \theta = 2 \qquad 0 \leqslant \theta \leqslant 2\pi$
๐ Reveal Solution
Use $\cos 2\theta = 1 - 2\sin^2\theta$ (the form that produces only $\sin\theta$ terms):
$1 - 2\sin^2\theta + 3\sin\theta = 2$
$-2\sin^2\theta + 3\sin\theta + 1 - 2 = 0$
$-2\sin^2\theta + 3\sin\theta - 1 = 0$
$2\sin^2\theta - 3\sin\theta + 1 = 0$
$(2\sin\theta - 1)(\sin\theta - 1) = 0$
$\sin\theta = \dfrac{1}{2}$ or $\sin\theta = 1$
For $\sin\theta = \dfrac{1}{2}$: $\theta = \dfrac{\pi}{6}$ and $\theta = \pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6}$.
For $\sin\theta = 1$: $\theta = \dfrac{\pi}{2}$.
All three values lie in $[0, 2\pi]$.
$\boxed{\theta = \dfrac{\pi}{6},\; \dfrac{\pi}{2},\; \dfrac{5\pi}{6}}$
Attempt each question on paper first, then click Reveal Solution to check your work. Questions are ordered from easiest to hardest.
Solve $\tan^2 \theta - 3 \tan \theta - 4 = 0 \qquad 0^\circ \leqslant \theta \leqslant 180^\circ$
๐ก Hint
This is a quadratic in $\tan\theta$ (Strategy 2). Let $u = \tan\theta$ and factorise $u^2 - 3u - 4 = 0$.
Remember: $\tan\theta$ is negative in Q2 (between $90^{\circ}$ and $180^{\circ}$), so a negative root gives a valid solution in this interval.
๐ Reveal Solution
$\tan^2\theta - 3\tan\theta - 4 = 0$
Find two numbers that multiply to $-4$ and add to $-3$: these are $-4$ and $+1$.
$(\tan\theta - 4)(\tan\theta + 1) = 0$
So $\tan\theta = 4$ or $\tan\theta = -1$.
Principal value: $\theta = \arctan(4) \approx 75.96^{\circ}$ (Q1).
Adding $180^{\circ}$: $\theta \approx 255.96^{\circ}$ (out of range).
Principal value: $\theta = \arctan(-1) = -45^{\circ}$ (out of range).
Add $180^{\circ}$: $\theta = -45^{\circ} + 180^{\circ} = 135^{\circ}$ (Q2, in range).
$\boxed{\theta \approx 75.96^{\circ},\; 135^{\circ}}$ (2 d.p. for the first; the second is exact.)
Solve $\cos 3 \theta = 1 - 3\cos \theta$ for $0^\circ \leqslant \theta \leqslant 360^\circ$
Hint: write $\cos 3\theta$ as $\cos(2\theta + \theta)$.
๐ก Hint
Use the compound angle formula: $\cos(2\theta + \theta) = \cos 2\theta\cos\theta - \sin 2\theta\sin\theta$.
Then substitute $\cos 2\theta = 2\cos^2\theta - 1$ and $\sin 2\theta = 2\sin\theta\cos\theta$. After simplifying, you should get a cubic equation in $\cos\theta$ that factorises.
๐ Reveal Solution
$\cos 3\theta = \cos 2\theta \cos\theta - \sin 2\theta \sin\theta$
Substitute $\cos 2\theta = 2\cos^2\theta - 1$ and $\sin 2\theta = 2\sin\theta\cos\theta$:
$\cos 3\theta = (2\cos^2\theta - 1)\cos\theta - 2\sin\theta\cos\theta \cdot \sin\theta$
$= 2\cos^3\theta - \cos\theta - 2\sin^2\theta\cos\theta$
$= 2\cos^3\theta - \cos\theta - 2(1 - \cos^2\theta)\cos\theta$
$= 2\cos^3\theta - \cos\theta - 2\cos\theta + 2\cos^3\theta$
$= 4\cos^3\theta - 3\cos\theta$
$4\cos^3\theta - 3\cos\theta = 1 - 3\cos\theta$
$4\cos^3\theta = 1$
$\cos^3\theta = \dfrac{1}{4}$
$\cos\theta = \sqrt[3]{\dfrac{1}{4}} \approx 0.62996$
Principal value: $\theta = \arccos(0.62996) \approx 50.95^{\circ}$ (Q1).
Second solution: $\theta = 360^{\circ} - 50.95^{\circ} = 309.05^{\circ}$ (Q4).
$\boxed{\theta \approx 50.95^{\circ},\; 309.05^{\circ}}$ (2 d.p.)
Solve $\cos \theta + 3 \sin \theta = 2 \qquad 0^\circ \leqslant \theta \leqslant 360^\circ$
๐ก Hint
This is a linear combination of $\cos\theta$ and $\sin\theta$ โ use Strategy 3 (R-form).
Here $a = 1$ (coefficient of $\cos\theta$) and $b = 3$ (coefficient of $\sin\theta$). Compute $R = \sqrt{a^2 + b^2}$ and $\tan\alpha = b/a$, then solve $R\cos(\theta - \alpha) = 2$.
๐ Reveal Solution
$a = 1$, $b = 3$. $R = \sqrt{1^2 + 3^2} = \sqrt{10}$.
$\tan\alpha = \dfrac{b}{a} = \dfrac{3}{1} = 3 \Rightarrow \alpha = \arctan(3) \approx 71.57^{\circ}$.
So $\cos\theta + 3\sin\theta = \sqrt{10}\cos(\theta - 71.57^{\circ})$.
$\sqrt{10}\cos(\theta - 71.57^{\circ}) = 2$
$\cos(\theta - 71.57^{\circ}) = \dfrac{2}{\sqrt{10}} \approx 0.6325$
Let $u = \theta - 71.57^{\circ}$. When $0^{\circ} \leq \theta \leq 360^{\circ}$, we have $-71.57^{\circ} \leq u \leq 288.43^{\circ}$.
$u = \arccos(0.6325) \approx 50.77^{\circ}$ (Q1), or $u = -50.77^{\circ}$ (Q4, since $\cos$ is even).
Both lie in the interval $[-71.57^{\circ}, 288.43^{\circ}]$.
$\theta = u + 71.57^{\circ}$.
$\theta = 50.77^{\circ} + 71.57^{\circ} = 122.34^{\circ}$.
$\theta = -50.77^{\circ} + 71.57^{\circ} = 20.80^{\circ}$.
$\boxed{\theta \approx 20.80^{\circ},\; 122.34^{\circ}}$ (2 d.p.)
Solve $2\cos\left( \theta - \frac{\pi}{6}\right) = 1 \qquad -\pi \leqslant \theta \leqslant \pi$
๐ก Hint
The argument is already a single cosine of a linear expression in $\theta$ โ you can solve directly without expanding.
Divide by 2: $\cos\!\left(\theta - \dfrac{\pi}{6}\right) = \dfrac{1}{2}$. Then let $u = \theta - \dfrac{\pi}{6}$ and find the range of $u$ corresponding to $-\pi \leq \theta \leq \pi$.
๐ Reveal Solution
$\cos\!\left(\theta - \dfrac{\pi}{6}\right) = \dfrac{1}{2}$
Let $u = \theta - \dfrac{\pi}{6}$. When $-\pi \leq \theta \leq \pi$, we have $u \in \left[-\pi - \dfrac{\pi}{6},\; \pi - \dfrac{\pi}{6}\right] = \left[-\dfrac{7\pi}{6},\; \dfrac{5\pi}{6}\right]$.
Principal value: $u = \dfrac{\pi}{3}$ (in range).
Second solution in one period: $u = -\dfrac{\pi}{3}$ (in range, since $-\dfrac{\pi}{3} > -\dfrac{7\pi}{6}$).
Further solutions: $u = \dfrac{\pi}{3} + 2\pi = \dfrac{7\pi}{3}$ (out of range); $u = -\dfrac{\pi}{3} - 2\pi = -\dfrac{7\pi}{3}$ (out of range).
Also $u = \dfrac{\pi}{3} - 2\pi = -\dfrac{5\pi}{3}$ (out of range, since $-\dfrac{5\pi}{3} < -\dfrac{7\pi}{6}$); $u = -\dfrac{\pi}{3} + 2\pi = \dfrac{5\pi}{3}$ (out of range, since $\dfrac{5\pi}{3} > \dfrac{5\pi}{6}$).
$\theta = u + \dfrac{\pi}{6}$.
For $u = \dfrac{\pi}{3}$: $\theta = \dfrac{\pi}{3} + \dfrac{\pi}{6} = \dfrac{\pi}{2}$.
For $u = -\dfrac{\pi}{3}$: $\theta = -\dfrac{\pi}{3} + \dfrac{\pi}{6} = -\dfrac{\pi}{6}$.
Both lie in $[-\pi, \pi]$.
$\boxed{\theta = -\dfrac{\pi}{6},\; \dfrac{\pi}{2}}$
Solve $\cos(45^\circ - \theta) = 2 \sin(30^\circ + \theta) \qquad -180^\circ \leqslant \theta \leqslant 180^\circ$
๐ก Hint
Let $x = 30^{\circ} + \theta$. Then $45^{\circ} - \theta = 75^{\circ} - x$, so the equation becomes $\cos(75^{\circ} - x) = 2\sin x$.
Expand $\cos(75^{\circ} - x) = \cos 75^{\circ}\cos x + \sin 75^{\circ}\sin x$, collect the $\sin x$ terms, and form a $\tan x$ equation.
๐ Reveal Solution
Then $\theta = x - 30^{\circ}$ and $45^{\circ} - \theta = 45^{\circ} - (x - 30^{\circ}) = 75^{\circ} - x$.
The equation becomes $\cos(75^{\circ} - x) = 2\sin x$.
When $-180^{\circ} \leq \theta \leq 180^{\circ}$, we have $-150^{\circ} \leq x \leq 210^{\circ}$.
$\cos 75^{\circ}\cos x + \sin 75^{\circ}\sin x = 2\sin x$
$\cos 75^{\circ}\cos x = (2 - \sin 75^{\circ})\sin x$
$\tan x = \dfrac{\cos 75^{\circ}}{2 - \sin 75^{\circ}} = \dfrac{0.2588}{2 - 0.9659} = \dfrac{0.2588}{1.0341} \approx 0.2503$
Principal value: $x = \arctan(0.2503) \approx 14.06^{\circ}$ (in range).
Adding $180^{\circ}$: $x = 14.06^{\circ} + 180^{\circ} = 194.06^{\circ}$ (in range, since $194.06^{\circ} \leq 210^{\circ}$).
Subtracting $180^{\circ}$: $x = 14.06^{\circ} - 180^{\circ} = -165.94^{\circ}$ (in range, since $-165.94^{\circ} \geq -150^{\circ}$ โ no, $-165.94^{\circ} < -150^{\circ}$, so out of range).
$\theta = x - 30^{\circ}$.
For $x = 14.06^{\circ}$: $\theta = 14.06^{\circ} - 30^{\circ} = -15.94^{\circ}$.
For $x = 194.06^{\circ}$: $\theta = 194.06^{\circ} - 30^{\circ} = 164.06^{\circ}$.
Both lie in $[-180^{\circ}, 180^{\circ}]$.
$\boxed{\theta \approx -15.94^{\circ},\; 164.06^{\circ}}$ (2 d.p.)
Exit Ticket
Before you leave, tick the objectives you are confident you can achieve. Be honest โ this guides next lesson's review.