Pure · Trigonometric Modelling · Year 2

Solving Trigonometric Equations

Some equations mix a single angle with a double angle — the fix is always the same: use a double angle formula to rewrite everything in terms of one function, then solve as usual.

8 min read 3 worked examples

Why these equations are different

An equation like $\cos2\theta + 3\sin\theta = 2$ mixes a "$2\theta$" term with a "$\theta$" term — you can't collect these directly, because they're different angles. The fix is to substitute a double angle identity for $\cos2\theta$ (choosing the version written entirely in $\sin\theta$, to match the other term), which turns the whole equation into one function of a single angle.

Which version of cos 2θ to use

$\cos2\theta \equiv 1-2\sin^2\theta$ when the rest of the equation is in $\sin\theta$. $\cos2\theta\equiv2\cos^2\theta-1$ when the rest is in $\cos\theta$. Pick whichever keeps everything in one function.

Worked example 1

A double angle equation with two solution branches

Solve $\cos2\theta + 3\sin\theta = 2$ for $0 \le \theta < 360°$.

MethodSubstitute for cos 2θ, then factorise the resulting quadratic

Everything else is in $\sin\theta$, so use $\cos2\theta = 1-2\sin^2\theta$: $1-2\sin^2\theta+3\sin\theta=2$.

Choosing this version keeps the whole equation in terms of sinθ only.

Rearrange: $-2\sin^2\theta+3\sin\theta-1=0 \Rightarrow 2\sin^2\theta-3\sin\theta+1=0$.

Multiply through by −1 to make the leading coefficient positive.

Let $s=\sin\theta$: $2s^2-3s+1=0 \Rightarrow (2s-1)(s-1)=0 \Rightarrow s=\dfrac12$ or $s=1$.

This is now an ordinary quadratic in s — factorise as usual.

$\sin\theta=\dfrac12 \Rightarrow \theta=30°,150°$. $\sin\theta=1 \Rightarrow \theta=90°$.

Solve each linear trig equation separately over the given range.

▪ solved

Solutions: $\theta = 30°, 90°, 150°$.

Worked example 2

Recognising when to leave a factor alone

Solve $\cos2\theta = \sin\theta$ for $0 \le \theta < 360°$.

MethodSame substitution, but check every root — including the negative one

Substitute $\cos2\theta=1-2\sin^2\theta$: $1-2\sin^2\theta=\sin\theta$.

Both sides are now in terms of sinθ alone.

Rearrange to standard form: $2\sin^2\theta+\sin\theta-1=0$.

Move every term to one side.

Let $s=\sin\theta$: $2s^2+s-1=0 \Rightarrow (2s-1)(s+1)=0 \Rightarrow s=\dfrac12$ or $s=-1$.

Factorise the quadratic in s.

$\sin\theta=\dfrac12 \Rightarrow \theta=30°,150°$. $\sin\theta=-1 \Rightarrow \theta=270°$.

Don't discard the negative root — sine genuinely does equal −1 somewhere in the range.

▪ solved

Solutions: $\theta = 30°, 150°, 270°$.

Worked example 3 — Challenge

An equation that factorises without expanding cos 2θ

Solve $\sin2\theta = \cos\theta$ for $0 \le \theta < 360°$.

MethodRewrite sin 2θ, then factor out the common cos θ

Substitute $\sin2\theta=2\sin\theta\cos\theta$: $2\sin\theta\cos\theta = \cos\theta$.

This is the only double angle identity for sin 2θ, so there's no choice of version here.

Rearrange: $2\sin\theta\cos\theta-\cos\theta=0 \Rightarrow \cos\theta(2\sin\theta-1)=0$.

Do NOT divide both sides by cos θ — that would silently delete every solution where cos θ = 0.

$\cos\theta=0 \Rightarrow \theta=90°,270°$. $\sin\theta=\dfrac12 \Rightarrow \theta=30°,150°$.

Solve each factor separately, exactly as with any factorised equation.

▪ solved

Solutions: $\theta = 30°, 90°, 150°, 270°$.

Common mistake

Dividing an equation like $\cos\theta(2\sin\theta-1)=0$ through by $\cos\theta$ is one of the most common ways marks are lost — it treats $\cos\theta=0$ as impossible and throws away real solutions. Whenever a common factor appears, factorise and solve each bracket separately instead of dividing it out.

Check yourself

Solve $\cos2\theta + \cos\theta = 0$ for $0 \le \theta < 360°$.
(Hint: this time everything should end up in terms of cos θ — use the version of cos 2θ written with cos²θ.)

Method

Substitute $\cos2\theta=2\cos^2\theta-1$: $2\cos^2\theta-1+\cos\theta=0$.

This version keeps everything in terms of cos θ, matching the other term.

Let $c=\cos\theta$: $2c^2+c-1=0 \Rightarrow (2c-1)(c+1)=0 \Rightarrow c=\dfrac12$ or $c=-1$.

Factorise the resulting quadratic in c.

$\cos\theta=\dfrac12 \Rightarrow \theta=60°,300°$. $\cos\theta=-1 \Rightarrow \theta=180°$.

Solve each linear trig equation over the given range.

▪ solved

Solutions: $\theta = 60°, 180°, 300°$.

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Double Angle Formulae

R cos/sin Form