Pre-Calculus Trigonometric functions

Solving trig equations

Q: What period do you add for the general solution?

2 pi n for sine and cosine, and pi n for tangent, since tangent repeats every pi.

20 practice questions 0 video lessons Theory + worked examples

Solving Trigonometric Equations

Texas Precalculus (TEKS) • Standard P.5(N) • Trigonometric Functions

Solving Trigonometric Equations is a topic in Trigonometric Functions in the Texas Essential Knowledge and Skills (Precalculus, §111.42). It is aligned to Standard P.5(N), which requires students to generate and solve trigonometric equations.

Solving a trigonometric equation uses reference angles and the quadrants to find solutions on one period, then adds the period for the general solution.

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Practice questions

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Theory

A trigonometric equation is solved by finding every angle that satisfies it. Because trig functions repeat, there are usually infinitely many solutions, described by a general formula.

The plan:

Isolate the trig function.
Reference angle gives the size of the solution.
Quadrants (from the sign) give which solutions occur in one period.
Add the period (\(2\pi n\) for sine/cosine, \(\pi n\) for tangent) for the general solution.

Read the interval. \([0,2\pi)\) wants the solutions in one turn; “all solutions” wants the general formula with \(n\).

On the unit circle, \(\sin x=\dfrac12\) at the reference angle \(\dfrac{\pi}{6}\) (and its QII partner).

Graphically, solutions are where \(y=\sin x\) meets the line \(y=\dfrac12\).

General solutions add the period:

\[\sin x=\sin\alpha:\ x=\alpha+2\pi n\ \text{or}\ \pi-\alpha+2\pi n;\qquad \tan x=\tan\alpha:\ x=\alpha+\pi n\]

sine solutions are alpha plus 2 pi n or pi minus alpha plus 2 pi n; tangent solutions are alpha plus pi n

Sine and cosine repeat every \(2\pi\); tangent every \(\pi\). Match the period to the function.

How to solve a trig equation

Isolate the trig function (and factor if it appears to a power).
Reference angle from the magnitude of the value.
Place solutions in the correct quadrants over one period.
Generalize by adding the period, or restrict to the given interval.

Example 1 — Basic equation on one period

Solve \(\sin x=\dfrac12\) on \([0,2\pi)\).

Solution

The reference angle is \(\dfrac{\pi}{6}\); sine is positive in QI and QII.

\(x\)	\(=\)	\(\dfrac{\pi}{6}\)
\(x\)	\(=\)	\(\pi-\dfrac{\pi}{6}=\dfrac{5\pi}{6}\)

solutions are pi over 6 and five pi over six

Example 2 — Isolate first

Solve \(2\cos x-1=0\) on \([0,2\pi)\).

Solution

Isolate \(\cos x\), then use the reference angle.

\(\cos x\)	\(=\)	\(\dfrac12\)
\(x\)	\(=\)	\(\dfrac{\pi}{3},\ \dfrac{5\pi}{3}\)

(cosine is positive in QI and QIV).

solutions are pi over 3 and five pi over 3

Example 3 — Factor a trig equation

Solve \(2\sin^2 x-\sin x=0\) on \([0,2\pi)\).

Solution

Factor out \(\sin x\).

\(\sin x(2\sin x-1)\)	\(=\)	\(0\)
\(\sin x=0\)	\(\Rightarrow\)	\(x=0,\ \pi\)
\(\sin x=\dfrac12\)	\(\Rightarrow\)	\(x=\dfrac{\pi}{6},\ \dfrac{5\pi}{6}\)

solutions are 0, pi, pi over 6, and five pi over six

Example 4 — General solution

Give all solutions of \(\tan x=1\).

Solution

Tangent has period \(\pi\), so add integer multiples of \(\pi\) to the principal solution.

\(x\)

\(=\)

\(\dfrac{\pi}{4}+\pi n,\ n\in\mathbb{Z}\)

all solutions are pi over 4 plus multiples of pi

Common pitfalls

Don't lose solutions. Each period usually gives two (one per quadrant with the right sign); check both.

Isolate before factoring. Never divide by \(\sin x\) or \(\cos x\) — you'd drop the solutions where it is zero.

Match the period. Add \(2\pi n\) for sine/cosine but \(\pi n\) for tangent.

Frequently asked questions

How do you solve a trigonometric equation?

Isolate the trig function, use the reference angle for the size, place solutions in the correct quadrants, and add the period for the general solution.

Why are there infinitely many solutions?

Because trig functions are periodic: once you find one solution, adding whole periods gives more. The general solution captures them all with an integer \(n\).

Why shouldn't you divide by sin x or cos x?

Dividing by a trig factor discards the solutions where it equals zero. Factor instead and set each factor to zero.

What period do you add for the general solution?

\(2\pi n\) for sine and cosine, and \(\pi n\) for tangent, since tangent repeats every \(\pi\).

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