How to Solve Trigonometric Equations?

A step-by-step guide to solving trigonometric equations

Trigonometric equations are similar to algebraic equations and can be linear equations, quadratic equations, or polynomial equations.

To solve trigonometric equations, we use some general results and solutions of trigonometric equations. These results are as follows:

• For any real numbers \(x\) and \(y\), \(sin\: x = sin\: y\) implies \(x=n\pi +\left(-1\right)^ny\), where \(n ∈ Z\).
• For any real numbers \(x\) and \(y\), \(cos\: x = cos\: y\) implies \(x = 2nπ ± y\), where \(n ∈ Z\).
• If \(x\) and \(y\) are not odd multiples of \(\frac{π}{2}\), then \(tan\: x = tan\: y\) implies \(x = nπ + y\), where \(n ∈ Z\).

Solving trigonometric equations

We have two types of solutions to the trigonometric equations:

• Principal Solution:

The initial values of angles for trigonometric functions are called principal solutions.

• General Solution: 

The values of the angles for the same answer of the trigonometric function are called the general solution of the trigonometric function. The general solutions of \(sin\:θ\), \(cos\:θ\), and \(tan\:θ\) are as follows.

• \(sin\:θ = sin\:α\), and the general solution is \(θ=n\pi +\left(-1\right)^nα\), where \(n ∈ Z\)
• \(cos\:θ = cos\:α\), and the general solution is \(θ = 2nπ + α\), where \(n ∈ Z\)
• \(tan\:θ = tan\:α\), and the general solution is \(θ = nπ + α\), where \(n ∈ Z\)

Steps to solve trigonometric equations

To solve a trigonometric equation, the following steps should be followed.

• Convert the given trigonometric equation to an equation with a single trigonometric ratio (sin, cos, tan).
• Change the equation with the trigonometric equation, having multiple angles, or submultiple angles into a simple angle.
• Now express the equation as a polynomial equation, a quadratic equation, or a linear equation.
• Solve the trigonometric equation similar to normal equations and find the value of the trigonometric ratio.
• The angle of the trigonometric ratio or the value of the trigonometric ratio shows the solution of the trigonometric equation.

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Solving Trigonometric Equations – Example 1:

Find the principal solutions of the equation \(tan x=-\frac{1}{\sqrt{3}}\).

Solution:

We know that \(tan\left(\frac{\pi }{6}\right)=\frac{1}{\sqrt{3}}\)

\(tan\:\left(\pi \:-\:\frac{\pi }{6}\:\right)=-tan\:\left(\frac{\pi }{6}\right)=-\:\frac{1}{\sqrt{3}}\)

\(tan\:\left(2\pi \:-\:\frac{\pi }{6}\right)=-tan\left(\frac{\pi }{6}\right)=-\:\frac{1}{\sqrt{3}}\)

So, the principal solutions are \(tan\: (π – \frac{π}{6}) = tan\: (\frac{5π}{6})\) and \(tan\: (2π – \frac{π}{6}) = tan\: (\frac{11π}{6})\).

Exercises for Solving Trigonometric Equations

Solve each trigonometric function for all possible values in degree.

1. \(\color{blue}{cos\:x+\sqrt{3}=-cos\:x}\)
2. \(\color{blue}{sin\:2x-\frac{\sqrt{3}}{2}=0}\)
3. \(\color{blue}{4\:sin\:\theta -1=2\:sin\:\theta +1}\)
4. \(\color{blue}{csc\:x+cot\:x=1}\)

1. \(\color{blue}{\:x=150^{\circ \:}+360^{\circ \:}n,\:x=210^{\circ \:}+360^{\circ \:}n}\)
2. \(\color{blue}{x=30^{\circ \:}+180^{\circ \:}n,\:x=60^{\circ \:}+180^{\circ \:}n}\)
3. \(\color{blue}{\:θ=90^{\circ \:}+360^{\circ \:}n}\)
4. \(\color{blue}{\:x=90^{\circ \:}+360^{\circ \:}n}\)

Recommended EffortlessMath Books

For a workbook that builds trigonometric equations into a full trig unit, the Trigonometry for Beginners walks through every solution technique with worked examples. For pre-calc-level coverage that prepares you for calculus, see the Pre-Calculus for Beginners.

Frequently Asked Questions

How do you solve a trigonometric equation?

Isolate the trig function with algebra. Use inverse trig or the unit circle to find solutions in \([0, 2\pi)\). Then add the period times an integer (\(2\pi k\) for sin/cos, \(\pi k\) for tan) to capture all solutions. Always verify by plugging answers back in.

What is the general solution of a trig equation?

The general solution includes all infinitely many angles that satisfy the equation. For \(\sin(x)=\frac{1}{2}\), the general solution is \(x=\frac{\pi}{6}+2\pi k\) or \(x=\frac{5\pi}{6}+2\pi k\) for any integer \(k\). The \(2\pi k\) accounts for sine’s period.

How do you solve sin(x) = 1/2?

Sine equals \(\frac{1}{2}\) at \(x=\frac{\pi}{6}\) and \(x=\frac{5\pi}{6}\) in \([0, 2\pi)\). General solution: \(x=\frac{\pi}{6}+2\pi k\) or \(x=\frac{5\pi}{6}+2\pi k\) for any integer \(k\). The unit circle is the fastest way to spot both quadrants.

How do you solve cos(x) = -1/2?

Cosine equals \(-\frac{1}{2}\) in quadrants II and III, at \(x=\frac{2\pi}{3}\) and \(x=\frac{4\pi}{3}\) in \([0, 2\pi)\). General solution: \(x=\frac{2\pi}{3}+2\pi k\) or \(x=\frac{4\pi}{3}+2\pi k\).

How do you solve tan(x) = 1?

Tangent equals 1 at \(x=\frac{\pi}{4}\). Since tangent has period \(\pi\) (shorter than sin/cos’s \(2\pi\)), the general solution is \(x=\frac{\pi}{4}+\pi k\) for any integer \(k\). One angle plus \(\pi k\) covers both quadrants tan visits per cycle.

How do you handle trig equations with multiple angles?

For \(\sin(2x)=\frac{1}{2}\), solve as if for sine, then divide. First, \(2x=\frac{\pi}{6}+2\pi k\) or \(2x=\frac{5\pi}{6}+2\pi k\). Divide each by 2: \(x=\frac{\pi}{12}+\pi k\) or \(x=\frac{5\pi}{12}+\pi k\). The period of \(\sin(2x)\) is \(\pi\), not \(2\pi\) — so more solutions fit in \([0, 2\pi)\).

How do you solve trig equations by factoring?

Move everything to one side, factor like a quadratic, then set each factor to 0. Example: \(2\sin^2(x)+\sin(x)-1=0\) factors as \((2\sin x-1)(\sin x+1)=0\). So \(\sin x=\frac{1}{2}\) (giving \(x=\frac{\pi}{6}, \frac{5\pi}{6}\)) or \(\sin x=-1\) (giving \(x=\frac{3\pi}{2}\)) within \([0, 2\pi)\).

What is an extraneous solution in trig?

An answer that satisfies a modified equation (after squaring or multiplying) but not the original. Squaring both sides can flip signs and create false roots. After you finish solving, plug every candidate back into the original equation. Discard anything that does not check.

What identities help solve trig equations?

Pythagorean identities (\(\sin^2 x+\cos^2 x=1\)), double-angle formulas (\(\sin 2x=2\sin x\cos x\), \(\cos 2x=1-2\sin^2 x\)), sum-to-product, and product-to-sum. Use them to rewrite a mixed-function equation in one trig function so you can isolate and solve.

How do you find all solutions in a given interval?

Write the general solution, then plug in \(k=0, 1, -1, 2, -2, \ldots\) and keep only the values that land in the requested interval. Example: in \([0, 2\pi)\) for \(\sin(x)=\frac{1}{2}\), the general solution \(x=\frac{\pi}{6}+2\pi k\) gives \(x=\frac{\pi}{6}\) (\(k=0\)); higher \(k\) overshoots the interval.

Related EffortlessMath Lessons

If a topic on this page feels rusty, these short lessons go deeper:

• The unit circle
• Trigonometric identities
• Inverse trigonometric functions
• How to graph sine and cosine
• Rules of exponents