Solving Second-Degree Trigonometric Equations

A second-degree trigonometric equation has the trig function squared, such as 2sin²x − sin x − 1 = 0. It's structured exactly like a quadratic equation, just with sin x (or cos x, tan x) standing in for the variable.

The cleanest approach is to substitute a variable, such as y = sin x, which turns the trig equation into a standard quadratic: 2y² − y − 1 = 0. This quadratic can be factored or solved with the quadratic formula just like any other.

Solve 2sin²x − sin x − 1 = 0 for 0° ≤ x < 360°. Substituting y = sin x gives 2y² − y − 1 = 0, which factors to (2y + 1)(y − 1) = 0. So y = −1/2 or y = 1, meaning sin x = −1/2 or sin x = 1.

Solving sin x = 1 gives x = 90°. Solving sin x = −1/2 (sine negative, in quadrants III and IV, reference angle 30°) gives x = 210° and x = 330°. The complete solution set is x = 90°, 210°, 330°.

Once the quadratic is factored, each factor becomes its own first-degree trig equation to solve using the reference-angle-and-quadrant method.

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• Trigonometry