Trigonometric equations: exact and numerical solutions

A trigonometric equation is true only for particular values of its variable. For example, $sin x=1/2$ is an equation, while $sin^2x+cos^2x\equiv1$ is an identity true for every real $x$.

The central difficulty is not finding one angle. It is finding every angle in the stated interval. In radians,

$$\sin x=\frac12,qquad 0\leq x<2\pi$$

has the two solutions $x=\pi/6$ and $x=5\pi/6$. A calculator’s inverse sine returns only the principal value $\pi/6$.

Prerequisites

You should be able to:

• use exact trigonometric values and signs in the four quadrants;
• interpret periods and transformations of trigonometric graphs;
• distinguish reciprocal functions from inverse trigonometric functions;
• rearrange equations, factorise quadratics and use trigonometric identities;
• convert between degrees and radians.

Always check the calculator’s angle mode. A degree answer from a radian calculation is not a small rounding error. It is a different angle.

The three basic equations

Let $n\in\mathbb Z$. If $\alpha$ is one solution, periodicity and symmetry give the complete families

$$\boxed{\sin x=\sin\alpha \iff x=n\pi+(-1)^n\alpha},$$ $$\boxed{\cos x=\cos\alpha \iff x=2n\pi\pm\alpha},$$ $$\boxed{\tan x=\tan\alpha \iff x=n\pi+\alpha}.$$

In degrees, replace $pi$ by $180^\circ$ and $2\pi$ by $360^\circ$.

These formulae are useful for unrestricted solutions. For a finite interval, many learners find this safer:

1. Rearrange to isolate one trigonometric function.
2. Find a reference angle or principal value.
3. Use the signs of the function to select the correct quadrants.
4. Add or subtract full periods until every angle in the interval has been listed.
5. Substitute back, especially if the equation was transformed or squared.

The signs by quadrant are

$$\begin{array}{c|cccc} \text{quadrant} & \mathrm I & \mathrm{II} & \mathrm{III} & \mathrm{IV}\\ \hline \sin x & + & + & - & -\\ \cos x & + & - & - & +\\ \tan x & + & - & + & - \end{array}$$

Worked example 1: an exact sine equation

Solve

$$2\sin x=\sqrt3,qquad 0\leq x<2\pi.$$

First isolate sine:

$$\sin x=\frac{\sqrt3}{2}.$$

The reference angle is $\pi/3$. Sine is positive in quadrants I and II, so

$$x=\frac\pi3 \quad\text{or}\quad x=\pi-\frac\pi3=\frac{2\pi}{3}.$$

Therefore

$$\boxed{x=\frac\pi3,\ \frac{2\pi}{3}}.$$

The endpoint $2\pi$ is excluded by $x<2\pi$. Read strict and inclusive inequalities carefully.

Worked example 2: a negative cosine in degrees

Solve

$$3\cos\theta+1=0,qquad 0^\circ\leq\theta\leq360^\circ,$$

giving answers to $1$ decimal place.

Rearrange:

$$\cos\theta=-\frac13.$$

The calculator principal value is

$$\cos^{-1}\left(-\frac13\right)=109.471\ldots^\circ.$$

Cosine is negative in quadrants II and III. The second solution is the reflection in the horizontal axis:

$$360^\circ-109.471\ldots^\circ=250.528\ldots^\circ.$$

Thus

$$\boxed{\theta=109.5^\circ,\ 250.5^\circ}.$$

An alternative is to use the acute reference angle $\cos^{-1}(1/3)$, then form $180^\circ-\alpha$ and $180^\circ+\alpha$. Both approaches are valid if signs and quadrants are handled consistently.

Worked example 3: tangent over a longer interval

Solve

$$\tan x=-1,qquad -\pi\leq x\leq2\pi.$$

One solution is $x=-\pi/4$. Tangent has period $\pi$, so all solutions are

$$x=-\frac\pi4+n\pi,qquad n\in\mathbb Z.$$

Choose integers that place $x$ in the interval:

$$\begin{array}{c|c} n & -\pi/4+n\pi\\ \hline 0 & -\pi/4\\ 1 & 3\pi/4\\ 2 & 7\pi/4 \end{array}$$

The neighbouring values for $n=-1$ and $n=3$ lie outside. Hence

$$\boxed{x=-\frac\pi4,\ \frac{3\pi}{4},\ \frac{7\pi}{4}}.$$

Do not use a $2\pi$ step for tangent. Its period is $pi$.

Principal values are not complete solution sets

Inverse trigonometric functions return values in restricted ranges:

$$\begin{array}{c|c} \text{function} & \text{principal range}\\ \hline \arcsin x & [-\pi/2,\pi/2]\\ \arccos x & [0,\pi]\\ \arctan x & (-\pi/2,\pi/2) \end{array}$$

This makes each inverse a function, but it means a calculator usually supplies only one starting value. The graph, quadrants and period supply the rest.

Inverse does not mean reciprocal

$\sin^{-1}(1/2)=\pi/6$ means $\arcsin(1/2)=\pi/6$. It does not mean $1/\sin(1/2)$. The reciprocal of sine is cosecant, $\csc x$.

Self check 1

1. Solve $\sin x=-1/2$ for $0\leq x<2\pi$.
2. Solve $\cos\theta=\sqrt2/2$ for $-180^\circ\leq\theta\leq360^\circ$.
3. Solve $\tan x=\sqrt3$ for $0\leq x\leq3\pi$.

Answers

1. Sine is negative in quadrants III and IV: $x=7\pi/6,11\pi/6$.
2. Cosine is positive at $\theta=\pm45^\circ+360n^\circ$: $\theta=-45^\circ,45^\circ,315^\circ$.
3. One solution is $\pi/3$ and tangent has period $\pi$: $x=\pi/3,4\pi/3,7\pi/3$.

Equations with a multiple angle

For an equation involving $\sin(kx)$, $\cos(kx)$ or $\tan(kx)$, temporarily let

$$u=kx.$$

Transform the interval for $x$ into the corresponding interval for $u$, solve fully there, and only then divide by $k$. If the $u$ interval is not widened, solutions are lost.

Worked example 4: transform the interval

Solve

$$\sin(2x)=\frac12,qquad 0\leq x\leq2\pi.$$

Let $u=2x$. Since $0\leq x\leq2\pi$,

$$0\leq u\leq4\pi.$$

Within this two-cycle interval, $\sin u=1/2$ at

$$u=\frac\pi6,\ \frac{5\pi}{6},\ \frac{13\pi}{6},\ \frac{17\pi}{6}.$$

Now divide every value by $2$:

$$\boxed{x=\frac\pi{12},\ \frac{5\pi}{12},\ \frac{13\pi}{12},\ \frac{17\pi}{12}}.$$

There are four solutions because $sin(2x)$ completes two cycles while $x$ travels from $0$ to $2\pi$.

Worked example 5: a horizontal translation

Solve

$$\cos(3x-20^\circ)=0,qquad 0^\circ\leq x\leq180^\circ.$$

Set $u=3x-20^\circ$. Transform both endpoints:

$$-20^\circ\leq u\leq520^\circ.$$

Cosine is zero when

$$u=90^\circ+180n^\circ.$$

The values in the transformed interval are

$$u=90^\circ,270^\circ,450^\circ.$$

Solve $3x-20^\circ=u$:

$$x=\frac{u+20^\circ}{3}.$$

Therefore

$$\boxed{x=\frac{110}3^\circ,\ \frac{290}3^\circ,\ \frac{470}3^\circ}.$$

These exact fractional-degree answers may be written as $36\tfrac23^\circ$, $96\tfrac23^\circ$ and $156\tfrac23^\circ$.

Equations that are quadratic in a trigonometric function

Treat $\sin x$, $\cos x$ or $\tan x$ as one algebraic quantity. Factorise or use the quadratic formula, reject impossible function values, then solve each remaining basic equation.

Worked example 6: factorise, then solve each branch

Solve

$$2\cos^2x-3\cos x+1=0,qquad 0\leq x<2\pi.$$

Let $c=\cos x$. Then

$$2c^2-3c+1=0,$$

so

$$(2c-1)(c-1)=0.$$

Therefore

$$\cos x=\frac12 \quad\text{or}\quad \cos x=1.$$

For $cos x=1/2$,

$$x=\frac\pi3,\ \frac{5\pi}{3}.$$

For $cos x=1$, the interval contains $x=0$. Although $2\pi$ gives the same point on the unit circle, it is excluded. Thus

$$\boxed{x=0,\ \frac\pi3,\ \frac{5\pi}{3}}.$$

Worked example 7: reject an impossible branch

Solve

$$3\sin^2x+2\sin x-1=0,qquad 0^\circ\leq x\leq360^\circ.$$

Factorise:

$$(3\sin x-1)(\sin x+1)=0.$$

Hence

$$\sin x=\frac13 \quad\text{or}\quad \sin x=-1.$$

For the first branch,

$$\alpha=\sin^{-1}\left(\frac13\right)=19.471\ldots^\circ.$$

Sine is positive in quadrants I and II, giving

$$x=19.471\ldots^\circ,quad 160.528\ldots^\circ.$$

The second branch gives $x=270^\circ$. Therefore, to $1$ decimal place,

$$\boxed{x=19.5^\circ,\ 160.5^\circ,\ 270.0^\circ}.$$

If factorisation had produced $sin x=4/3$, that entire branch would be rejected because $-1\leq\sin x\leq1$.

Use identities to expose a solvable form

An equation containing different trigonometric functions usually needs rewriting in terms of one function. Useful identities include

$$\sin^2x+\cos^2x\equiv1,$$ $$\tan x\equiv\frac{\sin x}{\cos x}, \qquad 1+\tan^2x\equiv\sec^2x.$$

Later, compound and double angle formulae and harmonic form extend the range of equations that can be solved exactly.

Worked example 8: convert to one function

Solve

$$2\sin^2x+3\cos x=3,qquad 0\leq x<2\pi.$$

Use $\sin^2x=1-\cos^2x$:

$$2(1-\cos^2x)+3\cos x=3.$$

Rearranging gives

$$2\cos^2x-3\cos x+1=0.$$

Factorise:

$$(2\cos x-1)(\cos x-1)=0.$$

Thus $\cos x=1/2$ or $\cos x=1$, so

$$\boxed{x=0,\ \frac\pi3,\ \frac{5\pi}{3}}.$$

Worked example 9: factor before dividing

Solve

$$\sin x=2\sin x\cos x,qquad 0\leq x\leq2\pi.$$

Bring everything to one side and factorise:

$$\sin x(1-2\cos x)=0.$$

Therefore

$$\sin x=0 \quad\text{or}\quad \cos x=\frac12.$$

The first branch gives $x=0,\pi,2\pi$. The second gives $x=\pi/3,5\pi/3$. Hence

$$\boxed{x=0,\ \frac\pi3,\ \pi,\ \frac{5\pi}{3},\ 2\pi}.$$

Dividing the original equation by $\sin x$ would silently discard the solutions where $\sin x=0$. Factor first unless you have proved that the proposed divisor cannot be zero.

Worked example 10: division with a domain check

Solve

$$\sin x=\cos x,qquad 0\leq x<2\pi.$$

If $\cos x=0$, then $\sin x=\pm1$, so such values do not satisfy the equation. Division by $\cos x$ is therefore safe:

$$\frac{\sin x}{\cos x}=1,$$

so

$$\tan x=1.$$

Tangent is positive in quadrants I and III:

$$\boxed{x=\frac\pi4,\ \frac{5\pi}{4}}.$$

Squaring can introduce extraneous solutions

If $a=b$, then $a^2=b^2$. The converse is not always true because $a^2=b^2$ also allows $a=-b$. Any solution obtained after squaring must be checked in the original equation.

Worked example 11: check after squaring

Solve

$$\sin x=\cos^2x,qquad 0\leq x<2\pi.$$

Use $\cos^2x=1-\sin^2x$:

$$\sin x=1-\sin^2x.$$

Let $s=\sin x$:

$$s^2+s-1=0.$$

The quadratic formula gives

$$s=\frac{-1\pm\sqrt5}{2}.$$

Since $(-1-\sqrt5)/2<-1$, that value is impossible for sine. Let

$$a=\frac{\sqrt5-1}{2}.$$

Then $0<a<1$, and sine is positive in quadrants I and II:

$$\boxed{x=\arcsin a,\ \pi-\arcsin a}.$$

Numerically, these are $x\approx0.666$ and $x\approx2.475$ radians. Both satisfy the original equation. No squaring was needed here, but the range check plays the same protective role against invalid algebraic roots.

Reciprocal equations and excluded values

Equations involving $\sec x$, $\csc x$ or $\cot x$ can often be converted to cosine, sine or tangent. Keep track of where the reciprocal expression is undefined.

Worked example 12: solve a secant equation

Solve

$$2\sec x+1=0,qquad 0\leq x<2\pi.$$

Rearrange:

$$\sec x=-\frac12.$$

But every real secant value satisfies $|\sec x|\geq1$. Equivalently, taking reciprocals would require $\cos x=-2$, which is impossible. Therefore

$$\boxed{\text{there are no real solutions}.}$$

Never assume that every rearranged trigonometric equation has a solution. The ranges of sine and cosine are $[-1,1]$, while the ranges of secant and cosecant exclude $(-1,1)$.

Numerical equations

Some equations cannot be reduced to standard exact forms, for example

$$x=2\cos x.$$

Rewrite as a root problem,

$$f(x)=x-2\cos x=0,$$

then use a graph, sign-change search and a numerical method. A calculator’s solver needs sensible starting values and may return only one root.

Worked example 13: locate and refine a root

Find the positive solution of

$$x=2\cos x$$

to $3$ decimal places.

Define $f(x)=x-2\cos x$. Then

$$f(1)=1-2\cos1\approx-0.081,$$ $$f(1.1)=1.1-2\cos1.1\approx0.193.$$

Because $f$ is continuous, the sign change shows that a root lies in $(1,1.1)$. Numerical refinement gives

$$x=1.029866\ldots,$$

so

$$\boxed{x=1.030}.$$

For $x>0$, the equation requires $x\leq2$. On $(0,2)$,

$$f'(x)=1+2\sin x>0,$$

so $f$ is strictly increasing there. The positive root is therefore unique. A graph suggests this; the derivative proves it.

A reliable exam workflow

When solving a trigonometric equation in an interval:

1. Record whether angles are in degrees or radians.
2. Simplify algebraically and use identities to obtain one trigonometric function where possible.
3. Factorise rather than divide by an expression that might be zero.
4. Transform the interval when the angle is $kx+c$.
5. Find a principal value or exact reference angle.
6. Use quadrants, symmetry and period to list every candidate.
7. Return from the transformed angle to the original variable.
8. Reject out-of-range algebraic values and candidates outside the interval.
9. Check candidates in the original equation if you squared, multiplied by a variable expression or used a numerical method.
10. Round only at the end and state the requested accuracy.

Common misconceptions

• Stopping at the inverse value: $\arcsin a$ is generally one solution, not the solution set.
• Using the wrong period: sine and cosine repeat every $2\pi$; tangent repeats every $\pi$.
• Ignoring the transformed interval: if $u=3x$, the $u$ interval is three times as wide.
• Dividing away roots: division by $\sin x$ loses every solution for which $\sin x=0$.
• Accepting impossible values: $\sin x=1.4$ and $\cos x=-3$ have no real solutions.
• Confusing $\sin^2x$ with $\sin(x^2)$: $\sin^2x$ means $(\sin x)^2$.
• Mixing angle units: write the degree symbol on degree answers; radian answers have no unit symbol.
• Rounding a reference angle early: this can move a final answer across the required rounding boundary.

Self check 2

Solve each equation in the stated interval.

1. $2\cos^2x+\cos x-1=0$, for $0\leq x<2\pi$.
2. $\sin(3x)=0$, for $-\pi/3\leq x\leq\pi$.
3. $\tan(2\theta+15^\circ)=1$, for $0^\circ\leq\theta\leq180^\circ$.
4. $\sin^2x=\cos x$, for $0\leq x<2\pi$.
5. Explain why dividing $\sin x\cos x=\sin x$ by $\sin x$ gives an incomplete solution set.

Answers

1. $(2\cos x-1)(\cos x+1)=0$, so $\cos x=1/2$ or $-1$. Hence $x=\pi/3,\pi,5\pi/3$.
2. Let $u=3x$, so $-\pi\leq u\leq3\pi$. Since $\sin u=0$ when $u=n\pi$, $u=-\pi,0,\pi,2\pi,3\pi$. Thus $x=-\pi/3,0,\pi/3,2\pi/3,\pi$.
3. Let $u=2\theta+15^\circ$, so $15^\circ\leq u\leq375^\circ$. Since $\tan u=1$, $u=45^\circ,225^\circ$. Thus $\theta=15^\circ,105^\circ$.
4. Use $\sin^2x=1-\cos^2x$. Then $\cos^2x+\cos x-1=0$, so $\cos x=(\sqrt5-1)/2$; the other quadratic root is below $-1$. Therefore $$x=\arccos\left(\frac{\sqrt5-1}{2}\right),\quad 2\pi-\arccos\left(\frac{\sqrt5-1}{2}\right).$$
5. Division assumes $\sin x\ne0$ and loses the branch $\sin x=0$. Factoring gives $\sin x(\cos x-1)=0$, preserving all roots.

Next steps

Strengthen the algebra behind this lesson with trigonometric identities. Then learn to solve equations involving $\sin(A\pm B)$, $\cos2x$ and $\tan2x$ in compound and double angle formulae. For expressions of the form $a\sin x+b\cos x$, continue to harmonic form, and apply periodic solutions in trigonometric proof and modelling.