Mathematics 9709 · Pure Mathematics 1

Solving trigonometric equations

The calculator gives you one solution. The exam asks for every solution in the given interval.
The exam is checking that you find them all. The symmetry of the graphs gives you the other ones.

Get to $\text{(one ratio)}=k$ . Take the calculator's principal value. Then use the symmetry of the graph to find its partner. Then use the period to find all the rest.

The method that always works

No matter what the equation is, the steps are the same:

Rearrange (or use an identity) until you have one ratio equal to a number, e.g. $\sin x=k$ .
The calculator gives the principal value — your first solution.
Find the partner in $[0,360^\circ]$ using the rule for that ratio (below).
Add or subtract the period repeatedly to collect every solution inside the interval.

The partner rules come straight from the symmetry of the graphs:

$\sin x=k$ : the second solution is $180^\circ-x$ ; period $360^\circ$ .
$\cos x=k$ : the second solution is $360^\circ-x$ (that is, $-x$ ); period $360^\circ$ .
$\tan x=k$ : the solutions just step by the period $180^\circ$ .

cos x = k: the horizontal line cuts the cosine curve symmetrically, giving x and 360° − x.

The most common type: really a quadratic

The most common equation is really a quadratic.
How to spot it: you see a squared ratio mixed with a plain one, or a fraction to clear. Change it to one ratio. Then it is just GCSE algebra.

Worked example

Express $3\cos\theta = 8\tan\theta$ as a quadratic in $\sin\theta$ , then find the acute angle

This is 9709/12 Jun 2020 Q2. Use $\tan\theta=\tfrac{\sin\theta}{\cos\theta}$ . Then clear the fraction by multiplying by $\cos\theta$ :

$3\cos\theta=\dfrac{8\sin\theta}{\cos\theta}\;\Rightarrow\;3\cos^2\theta=8\sin\theta$ .
Swap to one ratio with $\cos^2\theta=1-\sin^2\theta$ : $3(1-\sin^2\theta)=8\sin\theta$ .
Rearrange to a standard quadratic in $\sin\theta$ .

Answer

3\sin^2\theta+8\sin\theta-3=0

Factorise $(3\sin\theta-1)(\sin\theta+3)=0$ . Throw out $\sin\theta=-3$ (it cannot happen, sine stays inside $[-1,1]$ ). So $\sin\theta=\tfrac13$ :

Answer

\theta=\sin^{-1}\big(\tfrac13\big)=19.5^\circ

Worked example

Solve $2\cos^2 x + \sin x = 2$ for $0 \le x \le 360^\circ$

One term is squared, one is not. Change the squared one with $\cos^2x=1-\sin^2x$ :

$2(1-\sin^2x)+\sin x=2 \;\Rightarrow\; \sin x-2\sin^2x=0$ .
Factorise, don't divide (dividing by $\sin x$ would lose solutions): $\sin x(1-2\sin x)=0$ .
Take each factor: $\sin x=0$ gives $x=0^\circ,180^\circ,360^\circ$ ; $\sin x=\tfrac12$ gives $x=30^\circ,150^\circ$ .

Answer

x=0^\circ,30^\circ,150^\circ,180^\circ,360^\circ

The 2x trap: make the interval bigger first

When the angle is $2x$ (or $2x+30^\circ$ ), you must make the interval bigger first.
If $0\le x\le360^\circ$ , then $0\le 2x\le720^\circ$ . Solve for $2x$ over that bigger range. Find all the values. Then halve them at the end.

See $2x$ ? Double the interval, solve, then halve. Halving first is the most costly mistake in this topic.

A harder one: an identity makes a quartic

Worked example

Show $3\tan^2\theta + 5\sin^2\theta \equiv \dfrac{8\sin^2\theta - 5\sin^4\theta}{1 - \sin^2\theta}$ , then solve ${}=9$ for $0^\circ \le \theta \le 270^\circ$

This is 9709/12 Mar 2025 Q7. Part (a) is all identity work. Part (b) then becomes a quadratic in $\sin^2\theta$ .

(a) Write $\tan^2\theta=\tfrac{\sin^2\theta}{\cos^2\theta}=\tfrac{\sin^2\theta}{1-\sin^2\theta}$ and put $5\sin^2\theta$ over the same denominator $1-\sin^2\theta$ ; the numerator tidies to $3\sin^2\theta+5\sin^2\theta(1-\sin^2\theta)=8\sin^2\theta-5\sin^4\theta$ .
(b) Set it equal to 9 and clear the denominator: $8\sin^2\theta-5\sin^4\theta=9(1-\sin^2\theta)$ , which rearranges to $5\sin^4\theta-17\sin^2\theta+9=0$ .
Treat as a quadratic in $\sin^2\theta$ : only $\sin^2\theta=\tfrac{17-\sqrt{109}}{10}$ is usable (the other root exceeds 1), giving $\sin\theta=\pm0.810$ .
Over $0\le\theta\le270^\circ$ : $\sin\theta=+0.810$ gives $54.1^\circ,125.9^\circ$ ; $\sin\theta=-0.810$ gives $234.1^\circ$ .

Answer

\theta=54.1^\circ,\ 125.9^\circ,\ 234.1^\circ

Where the marks go

Common mistake

Give every solution in the interval, not just the one from the calculator. Use the symmetry rule and the period to find the rest.

Common mistake

Never divide an equation by

\sin x

\cos x

. You lose the solutions where it equals zero. Factorise instead. (9709-style)

Common mistake

For a

2x

equation, double the interval before you solve, then halve at the end. Solving over

0

–

360^\circ

for

2x

loses half the answers.

Now you try

Solve $2\cos\theta=7-\dfrac{3}{\cos\theta}$ for $-90^\circ<\theta<90^\circ$ . (9709/12 Nov 2021 Q1)

Your turn— tap to reveal the worked answer (9709/12 Nov 2021 Q1)

Multiply everything by $\cos\theta$ to clear the fraction: $2\cos^2\theta=7\cos\theta-3$ , so $2\cos^2\theta-7\cos\theta+3=0$ .

Factorise $(2\cos\theta-1)(\cos\theta-3)=0$ . Throw out $\cos\theta=3$ (it cannot happen). Take $\cos\theta=\tfrac12$ . In $-90^\circ<\theta<90^\circ$ , cosine equals $\tfrac12$ at $\pm60^\circ$ :

Answer

\theta=-60^\circ,\ 60^\circ

PreviousTrigonometric identities NextThe binomial expansion

Solving trigonometric equations

The calculator gives you one solution. The exam asks for every solution in the given interval.
The exam is checking that you find them all. The symmetry of the graphs gives you the other ones.

Get to $\text{(one ratio)}=k$ . Take the calculator's principal value. Then use the symmetry of the graph to find its partner. Then use the period to find all the rest.

The method that always works

No matter what the equation is, the steps are the same:

Rearrange (or use an identity) until you have one ratio equal to a number, e.g. $\sin x=k$ .
The calculator gives the principal value — your first solution.
Find the partner in $[0,360^\circ]$ using the rule for that ratio (below).
Add or subtract the period repeatedly to collect every solution inside the interval.

The partner rules come straight from the symmetry of the graphs:

$\sin x=k$ : the second solution is $180^\circ-x$ ; period $360^\circ$ .
$\cos x=k$ : the second solution is $360^\circ-x$ (that is, $-x$ ); period $360^\circ$ .
$\tan x=k$ : the solutions just step by the period $180^\circ$ .

cos x = k: the horizontal line cuts the cosine curve symmetrically, giving x and 360° − x.

The most common type: really a quadratic

The most common equation is really a quadratic.
How to spot it: you see a squared ratio mixed with a plain one, or a fraction to clear. Change it to one ratio. Then it is just GCSE algebra.

Worked example

Express $3\cos\theta = 8\tan\theta$ as a quadratic in $\sin\theta$ , then find the acute angle

This is 9709/12 Jun 2020 Q2. Use $\tan\theta=\tfrac{\sin\theta}{\cos\theta}$ . Then clear the fraction by multiplying by $\cos\theta$ :

$3\cos\theta=\dfrac{8\sin\theta}{\cos\theta}\;\Rightarrow\;3\cos^2\theta=8\sin\theta$ .
Swap to one ratio with $\cos^2\theta=1-\sin^2\theta$ : $3(1-\sin^2\theta)=8\sin\theta$ .
Rearrange to a standard quadratic in $\sin\theta$ .

Answer

3\sin^2\theta+8\sin\theta-3=0

Factorise $(3\sin\theta-1)(\sin\theta+3)=0$ . Throw out $\sin\theta=-3$ (it cannot happen, sine stays inside $[-1,1]$ ). So $\sin\theta=\tfrac13$ :

Answer

\theta=\sin^{-1}\big(\tfrac13\big)=19.5^\circ

Worked example

Solve $2\cos^2 x + \sin x = 2$ for $0 \le x \le 360^\circ$

One term is squared, one is not. Change the squared one with $\cos^2x=1-\sin^2x$ :

$2(1-\sin^2x)+\sin x=2 \;\Rightarrow\; \sin x-2\sin^2x=0$ .
Factorise, don't divide (dividing by $\sin x$ would lose solutions): $\sin x(1-2\sin x)=0$ .
Take each factor: $\sin x=0$ gives $x=0^\circ,180^\circ,360^\circ$ ; $\sin x=\tfrac12$ gives $x=30^\circ,150^\circ$ .

Answer

x=0^\circ,30^\circ,150^\circ,180^\circ,360^\circ

The 2x trap: make the interval bigger first

When the angle is $2x$ (or $2x+30^\circ$ ), you must make the interval bigger first.
If $0\le x\le360^\circ$ , then $0\le 2x\le720^\circ$ . Solve for $2x$ over that bigger range. Find all the values. Then halve them at the end.

See $2x$ ? Double the interval, solve, then halve. Halving first is the most costly mistake in this topic.

A harder one: an identity makes a quartic

Worked example

Show $3\tan^2\theta + 5\sin^2\theta \equiv \dfrac{8\sin^2\theta - 5\sin^4\theta}{1 - \sin^2\theta}$ , then solve ${}=9$ for $0^\circ \le \theta \le 270^\circ$

This is 9709/12 Mar 2025 Q7. Part (a) is all identity work. Part (b) then becomes a quadratic in $\sin^2\theta$ .

(a) Write $\tan^2\theta=\tfrac{\sin^2\theta}{\cos^2\theta}=\tfrac{\sin^2\theta}{1-\sin^2\theta}$ and put $5\sin^2\theta$ over the same denominator $1-\sin^2\theta$ ; the numerator tidies to $3\sin^2\theta+5\sin^2\theta(1-\sin^2\theta)=8\sin^2\theta-5\sin^4\theta$ .
(b) Set it equal to 9 and clear the denominator: $8\sin^2\theta-5\sin^4\theta=9(1-\sin^2\theta)$ , which rearranges to $5\sin^4\theta-17\sin^2\theta+9=0$ .
Treat as a quadratic in $\sin^2\theta$ : only $\sin^2\theta=\tfrac{17-\sqrt{109}}{10}$ is usable (the other root exceeds 1), giving $\sin\theta=\pm0.810$ .
Over $0\le\theta\le270^\circ$ : $\sin\theta=+0.810$ gives $54.1^\circ,125.9^\circ$ ; $\sin\theta=-0.810$ gives $234.1^\circ$ .

Answer

\theta=54.1^\circ,\ 125.9^\circ,\ 234.1^\circ

Where the marks go

Common mistake

Give every solution in the interval, not just the one from the calculator. Use the symmetry rule and the period to find the rest.

Common mistake

Never divide an equation by

\sin x

\cos x

. You lose the solutions where it equals zero. Factorise instead. (9709-style)

Common mistake

For a

2x

equation, double the interval before you solve, then halve at the end. Solving over

0

–

360^\circ

for

2x

loses half the answers.

Now you try

Solve $2\cos\theta=7-\dfrac{3}{\cos\theta}$ for $-90^\circ<\theta<90^\circ$ . (9709/12 Nov 2021 Q1)

Your turn— tap to reveal the worked answer (9709/12 Nov 2021 Q1)

Multiply everything by $\cos\theta$ to clear the fraction: $2\cos^2\theta=7\cos\theta-3$ , so $2\cos^2\theta-7\cos\theta+3=0$ .

Factorise $(2\cos\theta-1)(\cos\theta-3)=0$ . Throw out $\cos\theta=3$ (it cannot happen). Take $\cos\theta=\tfrac12$ . In $-90^\circ<\theta<90^\circ$ , cosine equals $\tfrac12$ at $\pm60^\circ$ :

Answer

\theta=-60^\circ,\ 60^\circ