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Solving trigonometric equations involving multiple angles

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Trig Equations with Multiple Angles

A multiple-angle trig equation, like sin 2x equals one half, is solved by substituting u for the multiplied angle, solving over the correspondingly widened interval, and dividing the results by the multiplier. See a worked example showing why doubling the angle doubles the number of solutions.

Solving equations with a multiple angle

A multiple-angle trig equation, like sin 2x = ½, has the angle multiplied by a number (here, 2) instead of a plain x. Solving it takes one extra step compared to a single-angle equation: a substitution to keep track of the wider interval.

To solve sin 2x = ½ for 0° ≤ x < 360°, let u = 2x. Since x ranges up to 360°, u ranges up to 720° — twice as far, because doubling the angle means the full solution interval must double too.

Solving a trig equation with a multiple angle To solve sine of 2x equals one half for x between 0 and 360 degrees, substitute u for 2x so u ranges from 0 to 720 degrees. Solve sine u equals one half to get u equals 30, 150, 390, or 510 degrees. Dividing each by 2 gives x equals 15, 75, 195, or 255 degrees. Step 1: substitute sin 2x = ½, for 0° ≤ x < 360° let u = 2x, so 0° ≤ u < 720° Step 2: solve for u sin u = ½ → u = 30°, 150°, 390°, 510° Step 3: divide by 2 x = u ÷ 2 = 15°, 75°, 195°, 255° Four solutions in one period, because doubling the angle doubles how many times the cycle repeats.
Substituting u = 2x, solving for u over its doubled interval, then dividing by 2 to recover x.

Within 0° ≤ u < 720°, sin u = ½ has four solutions: u = 30°, 150°, 390°, and 510° (30° and 150° from the first cycle, plus another 360° added to each for the second cycle). Dividing each by 2 gives x = 15°, 75°, 195°, and 255°.

A multiple-angle equation like sin 2x = ½ has twice as many solutions in the same interval as sin x = ½ would, because the angle 2x completes its cycle twice as fast. This builds on solving first-degree trig equations and uses the same reference angle ideas.

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