What do I do with an equation like 2x = k?

Solve for the argument 2x over an interval that is scaled to match (e.g. [0^, 720^ ] if x [0^, 360^ ]), list all solutions, then divide each by 2 to get x.

Why shouldn’t I divide both sides by x?

Because x might be zero, and dividing by it discards the solutions where x = 0. Instead, bring everything to one side and factorise so no solutions are lost.

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E5PureStretch

Solving Trigonometric Equations

A trig equation like $sin x = 0.5$ has infinitely many solutions, because the graph repeats forever. The skill is finding every solution inside a given interval — and your calculator only ever hands you one of them.

30 min Video by TLMaths Trigonometry

The big picture

This is where the whole chapter pays off. Because sine, cosine and tangent are periodic, a single equation has endlessly many solutions, and exams ask for all of them within a stated range. The reliable method is graphical: draw the curve, draw the line, and read off every crossing in the interval. Harder equations first need an identity to become solvable — a quadratic in $sin x$ , say — and the biggest danger is losing solutions by dividing carelessly. Getting a systematic method here is the difference between scoring one mark and all of them.

A-Level Maths E7-07: Solving Basic Trigonometric EquationsWatch the walkthrough, then work through the notes below.

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What you'll be able to do

Solve basic equations like $sin x = k$ over a given interval
Use graph symmetry (or CAST) to find every solution in range
Solve equations that require an identity first, including quadratics in a ratio
Handle equations with a transformed argument such as $sin (2 x)$ or $cos (x + 3 0^{\circ})$

One equation, many solutions

To solve $sin x = 0.5$ in $[0^{\circ}, 36 0^{\circ}]$ , the calculator gives the $principal value$ $x = 3 0^{\circ}$ . But the graph of $y = sin x$ crosses the line $y = 0.5$ twice in that interval, so there is a second solution. Sketch $y = sin x$ and $y = 0.5$ , and read off both crossings.

Symmetry gives the second one directly: $sin (18 0^{\circ} - x) = sin x$ , so $x = 18 0^{\circ} - 3 0^{\circ} = 15 0^{\circ}$ . Always find the principal value, then use the graph or symmetry to complete the set.

1Principal value:

x = cos^{- 1} (- 0.5) = 12 0^{\circ}

2Cosine is symmetric about

18 0^{\circ}

: the other solution is

36 0^{\circ} - 12 0^{\circ} = 24 0^{\circ}

Answer

x = 12 0^{\circ}

x = 24 0^{\circ}

Tip — Sketch the graph every time. It shows instantly how many solutions lie in the interval and roughly where — a check the calculator cannot give you.

Equations that need an identity first

When an equation mixes ratios, use an identity to reduce it to one ratio. A classic is a $quadratic$ in $sin x$ or $cos x$ : replace $sin^{2} x$ with $1 - cos^{2} x$ so everything is in $cos x$ , then treat it as a quadratic (factorise or use the formula), and finally solve each simple equation.

1Replace

sin^{2} x = 1 - cos^{2} x

2 (1 - cos^{2} x) + cos x - 1 = 0

2Simplify:

- 2 cos^{2} x + cos x + 1 = 0

, i.e.

2 cos^{2} x - cos x - 1 = 0

3Factorise:

(2 cos x + 1) (cos x - 1) = 0

, so

cos x = - \frac{1}{2}

cos x = 1

cos x = - \frac{1}{2} \Rightarrow x = 12 0^{\circ}, 24 0^{\circ}

;

cos x = 1 \Rightarrow x = 0^{\circ}, 36 0^{\circ}

Answer

x = 0^{\circ}, 12 0^{\circ}, 24 0^{\circ}, 36 0^{\circ}

Why this works

Turn the equation into a quadratic in a $single$ ratio before solving. Mixing $sin$ and $cos$ is what blocks you; an identity removes the block.

Transformed arguments — adjust the interval

For an equation like $sin (2 x) = 0.5$ or $cos (x + 3 0^{\circ}) = 0.4$ , solve for the whole bracket first, but $adjust the interval to match$ . If $x$ runs over $[0^{\circ}, 36 0^{\circ}]$ then $2 x$ runs over $[0^{\circ}, 72 0^{\circ}]$ — so you must find all solutions for $2 x$ across that wider range before dividing by 2. Missing this doubles-or-halves the number of answers.

1Let

u = 2 x

; the interval becomes

0^{\circ} \leq u \leq 72 0^{\circ}

2Solve

sin u = \frac{3}{2}

u = 6 0^{\circ}, 12 0^{\circ}, 42 0^{\circ}, 48 0^{\circ}

3Divide each by 2:

x = 3 0^{\circ}, 6 0^{\circ}, 21 0^{\circ}, 24 0^{\circ}

Answer

x = 3 0^{\circ}, 6 0^{\circ}, 21 0^{\circ}, 24 0^{\circ}

Tip — Widen the interval for the bracket first, list all solutions there, and only then transform back to $x$ . This is the single most-missed step in the topic.

Think like an examiner

Find every solution in the stated interval, not just the calculator’s principal value — use the graph or symmetry.
For $sin (n x)$ or $cos (x + a)$ , extend the interval to match the argument before solving, then convert back.
When an equation has $sin^{2}$ and $cos$ , use $sin^{2} x = 1 - cos^{2} x$ to make a quadratic in one ratio.
Never divide an equation by $sin x$ or $cos x$ — you would delete solutions. Factorise instead.

Common misconceptions

Giving only the calculator’s single answer.

Use graph symmetry to find all solutions in the interval.

Solving

sin 2 x = k

over

[0^{\circ}, 36 0^{\circ}]

without widening for

2 x

Let the argument range over

[0^{\circ}, 72 0^{\circ}]

, solve, then divide by 2.

Dividing

sin x cos x = sin x

sin x

Move everything to one side and factorise:

sin x (cos x - 1) = 0

, keeping the

sin x = 0

solutions.

Leaving a mix of

sin

and

cos

and trying to solve directly.

Use an identity to reduce to one ratio first.

Solving toolkit

sin (18 0^{\circ} - x) = sin x

Second sine solution.

cos (36 0^{\circ} - x) = cos x

Second cosine solution.

sin^{2} x = 1 - cos^{2} x

Reduce to one ratio.

argument n x \Rightarrow widen interval

Adjust range before dividing.

Stretch yourself

Solve $tan (x - 4 5^{\circ}) = 1$ for $0^{\circ} \leq x \leq 36 0^{\circ}$ .

Hint — Let $u = x - 4 5^{\circ}$ , so the interval for $u$ is $- 4 5^{\circ} \leq u \leq 31 5^{\circ}$ . Solve $tan u = 1$ (period $18 0^{\circ}$ ) across that range, then convert back.

Questions students ask

Why does a trig equation have more than one solution?

Because the trig graphs repeat. A horizontal line usually crosses the curve several times in any interval, and each crossing is a solution. The calculator returns only one of them (the principal value).

How do I find all the solutions in a range?

Find the principal value from the calculator, then use the graph’s symmetry (e.g.

sin (18 0^{\circ} - x) = sin x

) or the period to generate the rest, keeping only those inside the stated interval.

What do I do with an equation like

sin 2 x = k

Solve for the argument

2 x

over an interval that is scaled to match (e.g.

[0^{\circ}, 72 0^{\circ}]

x \in [0^{\circ}, 36 0^{\circ}]

), list all solutions, then divide each by 2 to get

x

Why shouldn’t I divide both sides by

sin x

Because

sin x

might be zero, and dividing by it discards the solutions where

sin x = 0

. Instead, bring everything to one side and factorise so no solutions are lost.

Key takeaways

Trig equations have many solutions because the graphs are periodic.
Find the principal value, then use graph symmetry to get all solutions in range.
Reduce mixed equations to one ratio with an identity — often a quadratic.
For $sin (n x)$ or shifted arguments, widen the interval first, then convert back.
Factorise rather than divide by a trig ratio, to avoid losing solutions.

How this fits the course

Build on

Trigonometric Ratios and Graphs

Leads to

Addition and Double-Angle Formulae

Test yourself

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