What if the argument has a multiplier, like $\\cos(2x)$?

Solve for the general solutions of the argument $2x$ first. Then divide the entire solution equation by 2. Finally, increment your integer parameter $k$ to generate all values until they exceed the given interval bounds for $x$.

How do I handle negative intervals?

Use the same reference angles, but subtract multiples of the period (rotate clockwise on the unit circle) to find the coterminal angles that fit within the negative bounds.

Can I square both sides to solve?

Yes, but squaring both sides of an equation often introduces extraneous solutions. If you use this method to utilize Pythagorean identities, you must check every final candidate in the original equation to eliminate false roots.

How to solve a trig equation on an interval

Solving a trigonometric equation on a restricted interval requires isolating the trigonometric function and identifying all corresponding angles within the given bounds.

This method applies when you need a finite set of specific solutions, rather than a general solution set parameterized by an integer.

The setup

Identify the equation, the variable to solve for, and the given domain interval (commonly $[0, 2\\pi)$ or $[-\\pi, \\pi]$ ). If the equation contains multiple distinct trigonometric functions, use standard trigonometric identities to rewrite the equation in terms of a single trigonometric function.

The steps

Isolate the trigonometric expression using standard algebraic techniques (factoring, combining like terms, or the quadratic formula).
Determine the reference angle by evaluating the inverse trigonometric function of the isolated value.
Identify all quadrants where the function takes the required sign to find the general angles within one full rotation.
Add or subtract multiples of the period to generate a complete set of candidate solutions.
Filter the candidates, discarding any values that fall outside the specified interval.

Checking the result

Substitute each found angle back into the original equation. Confirm that the left-hand side equals the right-hand side. Finally, verify that each solution strictly obeys the boundary conditions of the interval (pay close attention to open vs. closed brackets like ( versus [).

Common errors

A frequent error is dividing both sides of an equation by a trigonometric function (e.g., dividing by $\\sin(x)$ ), which permanently deletes valid solutions where that function equals zero; always factor instead. Another common mistake is neglecting the interval endpoints, such as including $2\\pi$ when the interval is defined as $[0, 2\\pi)$ .

Worked example

Solve $2\\sin^2(x) - \\sin(x) - 1 = 0$ on the interval $[0, 2\\pi)$ .

Treat the equation as a quadratic in terms of $\\sin(x)$ . Factor the expression: $(2\\sin(x) + 1)(\\sin(x) - 1) = 0$

Set each factor to zero to isolate the trigonometric functions: $\\sin(x) = -\\frac{1}{2} \\quad \ ext{or} \\quad \\sin(x) = 1$

Evaluate the first equation, $\\sin(x) = -\\frac{1}{2}$ . The sine function is negative in Quadrants III and IV. The reference angle is $\\pi/6$ . Find the angles in $[0, 2\\pi)$ : $x = \\pi + \\frac{\\pi}{6} = \\frac{7\\pi}{6}$ $x = 2\\pi - \\frac{\\pi}{6} = \\frac{11\\pi}{6}$

Evaluate the second equation, $\\sin(x) = 1$ . The only angle in $[0, 2\\pi)$ where sine equals 1 is at the positive y-axis: $x = \\frac{\\pi}{2}$

Compile the final solution set: $x \\in \\left\\{ \\frac{\\pi}{2}, \\frac{7\\pi}{6}, \\frac{11\\pi}{6} \\right\\}$

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References: OpenStax Precalculus, Chapter 7: Trigonometric Identities and Equations · Stewart Calculus, Appendix D: Trigonometry · Khan Academy: Solving trigonometric equations