Solve 2sin x = 1 for 0 x 360^. [2 marks]

Solve tan x = -1 for 0 x 360^. [2 marks]

SQA SQA Higher 2018-style practice: Solve the equation 3sin^2 x - cos x - 1 = 0 for 0 x < 360^.

Make the equation one ratio using sin^2 x = 1 - cos^2 x: 3(1 - cos^2 x) - cos x - 1 = 0 (1 mark). Expand and tidy: 3 - 3cos^2 x - cos x - 1 = 0, so -3cos^2 x - cos x + 2 = 0, and multiplying by -1 gives 3cos^2 x + cos x - 2 = 0 (1 mark). Factorise as a quadratic in cos x: (3cos x - 2)(cos x + 1) = 0, so cos x = 23 or cos x = -1 (1 mark). Solve each: cos x = 23 gives x = 48.2^ (first quadrant) and x = 360^ - 48.2^ = 311.8^ (fourth quadrant); cos x = -1 gives x = 180^ (2 marks). Markers reward the identity substitution, the factorised quadratic, and all solutions in range.

SQA SQA Higher 2022-style practice: Solve 2sin(2x) = sqrt(3) for 0 x < 2π, giving your answers in radians.

Isolate the ratio: sin(2x) = sqrt(3)2 (1 mark). Substitute u = 2x; as 0 x < 2π, the interval becomes 0 u < 4π (1 mark). sin u = sqrt(3)2 has acute angle π3, and sine is positive in the first and second quadrants, so within [0, 4π): u = π3, 2π3, 7π3, 8π3 (1 mark). Divide each by 2: x = π6, π3, 7π6, 4π3 (1 mark). Markers reward isolating the ratio, expanding the interval for 2x, and dividing back to recover x.

ScotlandMathsSyllabus dot point

How do you solve trigonometric equations across a full interval, and how do the identities help you reduce them to a solvable form?

Solving trigonometric equations in degrees and radians over a given interval, using the CAST diagram and the symmetry of the graphs, the trigonometric identities, and equations that reduce to a quadratic in a single trigonometric ratio.

A focused answer to the SQA Higher Mathematics trigonometric equations content, covering solving equations in degrees and radians over a given interval, the CAST diagram and graph symmetry, the trigonometric identities, and equations that reduce to a quadratic in one ratio.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
Finding every solution
The identities
Quadratic trigonometric equations
Examples in context
Try this

What this dot point is asking

The SQA wants you to solve trigonometric equations in degrees and radians over a stated interval, use the CAST diagram and graph symmetry to find every solution, apply the trigonometric identities, and solve equations that become a quadratic in one trigonometric ratio.

Finding every solution

The CAST diagram tells you which quadrants give a positive value: cosine is positive in the fourth, all ratios positive in the first, sine positive in the second, and tangent positive in the third. The reliable routine is to isolate the trigonometric ratio, find the acute angle from the size of the value, decide from the sign which quadrants contain a solution, and then write the angle in each of those quadrants within the interval. Always match the unit (degrees or radians) to the interval the question gives.

The identities

These identities let you rewrite a mixed equation so it contains only one ratio. The Pythagorean identity swaps $\sin^2$ for $\cos^2$ (or back), and the tangent identity lets you turn $\dfrac{\sin\theta}{\cos\theta}$ into $\tan\theta$ or split a tangent back into sine over cosine.

Solve a multiple-angle equation in degrees

Solve $\cos 3x = \dfrac{1}{2}$ for $0 \le x < 360^\circ$ .

Step 1: Substitute to work in the new variable

Let $u = 3x$ . Because $x$ ranges over $[0^\circ, 360^\circ)$ , the variable $u$ ranges over $[0^\circ, 1080^\circ)$ . Expanding the interval first is what prevents missing the extra solutions that appear because of the multiplier of 3.

Step 2: Find every value of $u$ in the expanded interval

The equation is now $\cos u = \dfrac{1}{2}$ . The acute angle where $\cos = \dfrac{1}{2}$ is $60^\circ$ , and cosine is positive in the first and fourth quadrants. In the first revolution: $u = 60^\circ, 300^\circ$ . Adding $360^\circ$ for the second and third revolutions:

u = 60^\circ,\ 300^\circ,\ 420^\circ,\ 660^\circ,\ 780^\circ,\ 1020^\circ

Step 3: Divide each $u$ by 3 to recover $x$

Divide every solution by 3 to undo the substitution $u = 3x$ :

x = \dfrac{u}{3} = 20^\circ,\ 100^\circ,\ 140^\circ,\ 220^\circ,\ 260^\circ,\ 340^\circ

Final answer: $x = 20^\circ, 100^\circ, 140^\circ, 220^\circ, 260^\circ, 340^\circ$ .

Quadratic trigonometric equations

When the equation has a squared ratio, treat that ratio as a single variable, factorise the quadratic, then solve each linear factor across the interval. If the equation mixes $\sin^2 x$ with $\cos x$ , replace $\sin^2 x$ by $1 - \cos^2 x$ first so everything is in one ratio.

Solve

2\cos^2 x - \cos x - 1 = 0

for

0 \le x \le 360^\circ

The equation is a quadratic in $\cos x$ that can be factorised and then solved for each factor across the interval.

Step 1: Factorise as a quadratic in $\cos x$

Treat $\cos x$ as a single unknown. The product of $2$ and $-1$ is $-2$ , and $(-2)(+1) = -2$ with $-2 + 1 = -1$ , giving the factorisation:

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

Setting each factor to zero gives $\cos x = -\dfrac{1}{2}$ or $\cos x = 1$ .

Step 2: Solve $\cos x = -\dfrac{1}{2}$

The acute angle satisfying $\cos \theta = \dfrac{1}{2}$ is $60^\circ$ . Because cosine is negative, the solutions lie in the second and third quadrants:

x = 180^\circ - 60^\circ = 120^\circ \quad \text{and} \quad x = 180^\circ + 60^\circ = 240^\circ

Step 3: Solve $\cos x = 1$

A cosine value of $1$ is the maximum; it occurs where the graph touches its peak. Within $[0^\circ, 360^\circ]$ :

x = 0^\circ \quad \text{and} \quad x = 360^\circ

Final answer: $x = 0^\circ, 120^\circ, 240^\circ, 360^\circ$ .

Examples in context

Trigonometric equations find when a periodic quantity hits a target. If a tide depth is modelled by $d = 4 + 3\sin(30t)^\circ$ for time $t$ hours, the moment the depth is $5.5$ m solves $3\sin(30t)^\circ = 1.5$ , so $\sin(30t)^\circ = 0.5$ and $30t = 30^\circ$ or $150^\circ$ , giving $t = 1$ or $t = 5$ hours. Setting the model equal to a value and solving over the relevant interval is exactly the harbour-master's calculation.

Try this

Q1. Solve $2\sin x = 1$ for $0 \le x \le 360^\circ$ . [2 marks]

Cue. $\sin x = \dfrac{1}{2}$ , so $x = 30^\circ$ or $x = 150^\circ$ .

Q2. Solve $\tan x = -1$ for $0 \le x \le 360^\circ$ . [2 marks]

Cue. Tangent is negative in the second and fourth quadrants, so $x = 135^\circ$ or $x = 315^\circ$ .

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SQA Higher 20185 marksSolve the equation

3\sin^2 x - \cos x - 1 = 0

for

0 \le x < 360^\circ

Show worked answer →

Make the equation one ratio using $\sin^2 x = 1 - \cos^2 x$ : $3(1 - \cos^2 x) - \cos x - 1 = 0$ (1 mark).

Expand and tidy: $3 - 3\cos^2 x - \cos x - 1 = 0$ , so $-3\cos^2 x - \cos x + 2 = 0$ , and multiplying by $-1$ gives $3\cos^2 x + \cos x - 2 = 0$ (1 mark).

Factorise as a quadratic in $\cos x$ : $(3\cos x - 2)(\cos x + 1) = 0$ , so $\cos x = \dfrac{2}{3}$ or $\cos x = -1$ (1 mark).

Solve each: $\cos x = \dfrac{2}{3}$ gives $x = 48.2^\circ$ (first quadrant) and $x = 360^\circ - 48.2^\circ = 311.8^\circ$ (fourth quadrant); $\cos x = -1$ gives $x = 180^\circ$ (2 marks). Markers reward the identity substitution, the factorised quadratic, and all solutions in range.

SQA Higher 20224 marksSolve

2\sin\left(2x\right) = \sqrt{3}

for

0 \le x < 2\pi

, giving your answers in radians.

Show worked answer →

Isolate the ratio: $\sin(2x) = \dfrac{\sqrt{3}}{2}$ (1 mark).

Substitute $u = 2x$ ; as $0 \le x < 2\pi$ , the interval becomes $0 \le u < 4\pi$ (1 mark).

$\sin u = \dfrac{\sqrt{3}}{2}$ has acute angle $\dfrac{\pi}{3}$ , and sine is positive in the first and second quadrants, so within $[0, 4\pi)$ : $u = \dfrac{\pi}{3}, \dfrac{2\pi}{3}, \dfrac{7\pi}{3}, \dfrac{8\pi}{3}$ (1 mark).

Divide each by $2$ : $x = \dfrac{\pi}{6}, \dfrac{\pi}{3}, \dfrac{7\pi}{6}, \dfrac{4\pi}{3}$ (1 mark). Markers reward isolating the ratio, expanding the interval for $2x$ , and dividing back to recover $x$ .

Related dot points

Sources & how we know this

SQA Higher Mathematics Course Specification — SQA (2018)