How do I solve an equation like sine of two t equals k?

Let u equal two t, expand the domain to zero to four pi, solve sine of u equals k there, then divide every solution by two.

How do I solve an equation like sine of two t equals k?

Let u equal two t, expand the domain to zero to four pi, solve sine of u equals k there, then divide every solution by two.

Year 11 Methods (Unit 1 & 2) Trigonometric Functions

Solution Of Trigonometric Equations

Q: How many solutions does a trig equation have on 0 to 2 pi?

Usually two for sine equals k or cosine equals k, one in each matching quadrant, and two for tangent equals k. A multiple angle such as sine of n t equals k gives two n solutions.

Q: What is a reference angle?

It is the acute angle whose ratio equals the size of k. You read it from the exact values, then place it in the correct quadrants using symmetry.

Q: How do I know which quadrants to use?

Use ASTC and the sign of k. Sine is positive in quadrants one and two, cosine is positive in quadrants one and four, and tangent is positive in quadrants one and three. A negative value uses the other pair.

Q: Do I work in radians or degrees?

In Year 11 Methods work in radians and give exact answers such as pi over six, unless the domain is stated in degrees.

20 practice questions 1 video lesson Theory + worked examples

Learn to solve trigonometric equations for Queensland Year 11 Mathematical Methods (QCAA). Given sine, cosine or tangent equal to a value over a stated domain, you find every angle that fits.

You will learn to find the reference angle, place solutions in the correct quadrants using ASTC and unit-circle symmetry, give exact answers in radians, and handle multiple-angle equations — an essential skill in the trigonometry unit of the QCAA course.

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Every question with a fully worked solution.

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Theory

In Year 11 Mathematical Methods (QCAA, Unit 1), a trigonometric equation such as \(\sin t=k\), \(\cos t=k\) or \(\tan t=k\) is solved over a stated domain (for example \([0,2\pi]\)). Find the reference angle from the exact value, then use the quadrant signs (ASTC) and the periodicity of the unit circle to list every solution in the domain. This page covers exact radian solutions by symmetry.

A trigonometric equation asks for every angle \(t\) whose sine, cosine or tangent equals a given number \(k\). Because these functions repeat, an equation usually has several solutions in a domain, so the domain must always be stated.

The reference angle is the acute angle whose ratio equals \(|k|\); read it from the exact values at multiples of \(\dfrac{\pi}{6}\) and \(\dfrac{\pi}{4}\). The signs of the ratios in each quadrant follow ASTC: All positive in Q1, only Sine in Q2, only Tangent in Q3, only Cosine in Q4.

Use the sign of \(k\) to choose the quadrants, then place the reference angle in each with the correct symmetry: \(\pi-\theta\), \(\pi+\theta\) or \(2\pi-\theta\).

Reference angle first, then quadrants. The size of \(k\) fixes the reference angle; the sign of \(k\) and the ratio decide which quadrants hold a solution.

\(y=\sin t\) meets \(y=\tfrac12\) twice per cycle — solutions come in symmetric pairs.

On the unit circle the same height \(\tfrac12\) is reached in Q1 and Q2.

Over \([0,2\pi]\) with reference angle \(\theta\) (where \(\sin\theta=|k|\), etc.):

\[\sin t=k\gt 0:\quad t=\theta,\ \pi-\theta\]

t = θ, π - θ

\[\cos t=k\lt 0:\quad t=\pi-\theta,\ \pi+\theta\]

t = π - θ, π + θ

\[\tan t=k\lt 0:\quad t=\pi-\theta,\ 2\pi-\theta\]

t = π - θ, 2 π - θ

Expanded domain rule: to solve \(\sin nt=k\) over \([0,2\pi]\), let \(u=nt\) and solve \(\sin u=k\) over \([0,2n\pi]\); you get \(2n\) solutions.

How to solve a trigonometric equation over a domain

Isolate the ratio so the equation reads \(\sin t=k\), \(\cos t=k\) or \(\tan t=k\) (divide or rearrange first if needed).
Reference angle: find the acute \(\theta\) with the ratio equal to \(|k|\) from the exact values.
Quadrants: use the sign of \(k\) and ASTC to choose the quadrants, then write the solution in each using \(\pi-\theta\), \(\pi+\theta\) or \(2\pi-\theta\).
Domain: add or subtract \(2\pi\) to include every solution the stated domain allows (and, for \(nt\), expand the domain first).

Example 1 — Sine equals a positive value

Solve \(\sin t=\dfrac12\) for \(t\in[0,2\pi]\).

Solution

Reference angle — the acute angle with sine \(\tfrac12\):

\(\sin\theta\)	\(=\)	\(\tfrac12\)
\(\theta\)	\(=\)	\(\dfrac{\pi}{6}\)

Quadrants — \(k\gt 0\), so sine is positive in Q1 and Q2:

\(t\)	\(=\)	\(\theta \ \text{ or } \ \pi-\theta\)
	\(=\)	\(\dfrac{\pi}{6}\ \text{ or } \ \pi-\dfrac{\pi}{6}\)
	\(=\)	\(\dfrac{\pi}{6}\ \text{ or } \ \dfrac{5\pi}{6}\)

\(t=\dfrac{\pi}{6},\ \dfrac{5\pi}{6}\).

t = \frac{π}{6}, \frac{5 π}{6}

Example 2 — Cosine equals a negative value

Solve \(\cos t=-\dfrac{\sqrt3}{2}\) for \(t\in[0,2\pi]\).

Solution

Reference angle — ignore the sign, use \(\tfrac{\sqrt3}{2}\):

\(\cos\theta\)	\(=\)	\(\dfrac{\sqrt3}{2}\)
\(\theta\)	\(=\)	\(\dfrac{\pi}{6}\)

Quadrants — \(k\lt 0\), so cosine is negative in Q2 and Q3:

\(t\)	\(=\)	\(\pi-\theta \ \text{ or } \ \pi+\theta\)
	\(=\)	\(\pi-\dfrac{\pi}{6}\ \text{ or }\ \pi+\dfrac{\pi}{6}\)
	\(=\)	\(\dfrac{5\pi}{6}\ \text{ or } \ \dfrac{7\pi}{6}\)

\(t=\dfrac{5\pi}{6},\ \dfrac{7\pi}{6}\).

t = \frac{5 π}{6}, \frac{7 π}{6}

Example 3 — Isolate the ratio first

Solve \(2\sin t-\sqrt3=0\) for \(t\in[0,2\pi]\).

Solution

Isolate — make sine the subject:

\(2\sin t\)	\(=\)	\(\sqrt3\)
\(\sin t\)	\(=\)	\(\dfrac{\sqrt3}{2}\)

Reference angle:

\(\sin\theta\)	\(=\)	\(\dfrac{\sqrt3}{2}\)
\(\theta\)	\(=\)	\(\dfrac{\pi}{3}\)

Quadrants — sine positive in Q1 and Q2:

\(t\)	\(=\)	\(\dfrac{\pi}{3}\ \text{ or }\ \pi-\dfrac{\pi}{3}\)
	\(=\)	\(\dfrac{\pi}{3}\ \text{ or }\ \dfrac{2\pi}{3}\)

\(t=\dfrac{\pi}{3},\ \dfrac{2\pi}{3}\).

t = \frac{π}{3}, \frac{2 π}{3}

Example 4 — A multiple angle

Solve \(\cos 2t=-\dfrac12\) for \(t\in[0,2\pi]\).

Solution

Substitute — let \(u=2t\); as \(t\) runs over \([0,2\pi]\), \(u\) runs over \([0,4\pi]\):

\(\cos u\)

\(=\)

\(-\tfrac12\)

Reference angle and quadrants (cosine negative in Q2, Q3):

\(\theta\)	\(=\)	\(\dfrac{\pi}{3}\)
\(u\)	\(=\)	\(\dfrac{2\pi}{3},\ \dfrac{4\pi}{3},\ \dfrac{8\pi}{3},\ \dfrac{10\pi}{3}\)

Back-substitute \(t=\dfrac{u}{2}\):

\(t\)

\(=\)

\(\dfrac{\pi}{3},\ \dfrac{2\pi}{3},\ \dfrac{4\pi}{3},\ \dfrac{5\pi}{3}\)

\(t=\dfrac{\pi}{3},\ \dfrac{2\pi}{3},\ \dfrac{4\pi}{3},\ \dfrac{5\pi}{3}\).

t = \frac{π}{3}, \frac{2 π}{3}, \frac{4 π}{3}, \frac{5 π}{3}

Common pitfalls

Giving only one solution. A trig equation over \([0,2\pi]\) usually has two or more solutions; use symmetry to find them all, not just the calculator value.

Wrong quadrants for the sign. The sign of \(k\) picks the quadrants. Positive sine is Q1/Q2; negative cosine is Q2/Q3 — check with ASTC.

Forgetting to expand the domain for \(nt\). For \(\sin 2t=k\) you must solve over \([0,4\pi]\) before halving, or you will lose solutions.

Not isolating the ratio. Rearrange \(2\sin t-\sqrt3=0\) to \(\sin t=\tfrac{\sqrt3}{2}\) before finding the reference angle.

Frequently asked questions

How many solutions does a trig equation have on 0 to 2 pi?

Usually two for \(\sin t=k\) or \(\cos t=k\) (one per matching quadrant), and two for \(\tan t=k\). A multiple angle \(\sin nt=k\) gives \(2n\) solutions.

What is a reference angle?

The acute angle \(\theta\) whose ratio equals \(|k|\). You read it from the exact values, then place it in the correct quadrants using symmetry.

How do I know which quadrants to use?

Use ASTC and the sign of \(k\): sine positive in Q1/Q2, cosine positive in Q1/Q4, tangent positive in Q1/Q3; a negative value flips to the other pair.

How do I solve an equation like sin 2t = k?

Let \(u=2t\), expand the domain to \([0,4\pi]\), solve \(\sin u=k\) there, then divide every solution by \(2\).

Do I work in radians or degrees?

In Year 11 Methods work in radians and give exact answers like \(\dfrac{\pi}{6}\), unless the domain is stated in degrees.

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Solution Of Trigonometric Equations

📖 Theory

How to solve a trigonometric equation over a domain

Common pitfalls

Frequently asked questions

Theory