Trigonometric Equations with a Multiple Angle

Name the whole angle

Solve sin 2x = ½ for $0^{\circ }\le x\le 360^{\circ }$0° ≤ x ≤ 360°. The angle inside the sine is not x but 2x, twice it. The first step is to give that whole angle a name of its own: let u = 2x. The equation becomes sin u = ½.

That is a simple equation, and you know how to solve it. But it is an equation in u, and the question asks for x. Two things change because of that: the range of angles to search, and the last step, where each value of u is turned back into a value of x.

Stretch the range first

The range is given for x: $0^{\circ }\le x\le 360^{\circ }$0° ≤ x ≤ 360°. Multiply every part of it by 2 to get the range for u: $0^{\circ }\le 2x\le 720^{\circ }$0° ≤ 2x ≤ 720°, so $0^{\circ }\le u\le 720^{\circ }$0° ≤ u ≤ 720°.

As x makes one full turn, u = 2x makes two full turns. Every solution of sin u = ½ between 0° and 720° gives a value of x between 0° and 360°, so all of them are needed.

Four solutions in two turns

Solve sin u = ½ over the first turn of u. The inverse sine gives u = 30°, and the sine pairing gives u = 180° − 30° = 150°.

The sine repeats every 360°, so add a full turn to each: 30° + 360° = 390° and 150° + 360° = 510°. Both are below 720°. One more turn would give 750° and 870°, which are past the end of the range. So u = 30°, 150°, 390° or 510°.

Now turn each value of u back into x. Since u = 2x, $x=u/2$x = u / 2: x = 15°, 75°, 195° or 255°. All four lie between 0° and 360°, as they must.

Check one of them: at x = 195°, 2x = 390°, and sin 390° = sin (390° − 360°) = sin 30° = ½.

The range for u runs from 0° to 720°. The four solutions of sin u = ½ are marked: 30° and 150° in the first turn, and 390° and 510°, one full turn later, in the second.

The level line at 0.5 cuts the circle at the arms 30° and 150°: the first turn. Turn n up to 1 and each arm makes one more full turn, landing at 390° and 510° on the wave: the other two solutions for u.

The same four answers on the graph of sin 2x

The graph of y = sin 2x shows the same thing in terms of x. Doubling the angle squeezes the wave across: it completes a cycle every 180°, so there are two full waves between 0° and 360°.

The horizontal line y = ½ cuts each of those waves twice, so it meets the graph four times between 0° and 360°: at x = 15°, 75°, 195° and 255°.

y = sin 2x from 0° to 360°, with one square across for every 45°. It makes two full waves, and the dashed line y = ½ crosses it at 15°, 75°, 195° and 255°.

A shift inside the bracket

Solve cos(2x − 30°) = ½ for $0^{\circ }\le x\le 180^{\circ }$0° ≤ x ≤ 180°. Here the whole angle is u = 2x − 30°, and its range is built from the range for x in the same order as the angle is built from x: double, then subtract 30°.

Doubling every part of $0^{\circ }\le x\le 180^{\circ }$0° ≤ x ≤ 180° gives $0^{\circ }\le 2x\le 360^{\circ }$0° ≤ 2x ≤ 360°. Subtracting 30° from every part gives $-30^{\circ }\le 2x-30^{\circ }\le 330^{\circ }$−30° ≤ 2x − 30° ≤ 330°. So u runs from −30° to 330°.

Solve, then undo the steps in reverse

Solve cos u = ½. The inverse cosine gives u = 60°, and the cosine pairing gives u = 360° − 60° = 300°. Both lie between −30° and 330°.

Check for others. Subtracting a full turn gives 60° − 360° = −300° and 300° − 360° = −60°; both are below −30°. Adding a full turn gives 420° and 660°, both above 330°. The angle −60° is the one most easily included by mistake: its cosine is ½, but it falls just outside the range. So u = 60° or u = 300°.

Now undo the steps that built u, last one first. Add 30°: 2x = 90° or 2x = 330°. Then halve: x = 45° or x = 165°.

Check: at x = 45°, 2x − 30° = 60°, and cos 60° = ½. At x = 165°, 2x − 30° = 300°, and cos 300° = ½.

The range for u runs from −30° to 330°. Of the angles with a cosine of ½, only 60° and 300° lie inside it; −60° and 420° fall just outside.

A half angle shrinks the range

A multiple can be less than 1. Solve $\cos \left(\frac{x}{2}\right)=\frac{1}{2}$cos(x/2) = ½ for $0^{\circ }\le x\le 360^{\circ }$0° ≤ x ≤ 360°. Let $u=\frac{x}{2}$u = x/2. Halving the range gives $0^{\circ }\le u\le 180^{\circ }$0° ≤ u ≤ 180°, which is only half a turn.

The solutions of cos u = ½ are 60° and 300°, but 300° is outside 0° to 180°. So u = 60° only, and x = 2 × 60° = 120°. Check: $\cos \left(\frac{120^{\circ }}{2}\right)=\cos 60^{\circ }=\frac{1}{2}$cos(120°/2) = cos 60° = ½.

Doubling the angle doubles how often the wave repeats, so it doubles the number of solutions. Halving the angle halves it.

Tangent, and a calculator case

Solve tan 2x = 1 for $0^{\circ }\le x\le 180^{\circ }$0° ≤ x ≤ 180°. Let u = 2x, so $0^{\circ }\le u\le 360^{\circ }$0° ≤ u ≤ 360°. The tangent repeats every 180°, so tan u = 1 gives u = 45° and u = 45° + 180° = 225°. Halving gives x = 22.5° or x = 112.5°.

Solve sin 3x = 0.4 for $0^{\circ }\le x\le 180^{\circ }$0° ≤ x ≤ 180°. Let u = 3x, so $0^{\circ }\le u\le 540^{\circ }$0° ≤ u ≤ 540°: one and a half turns. The inverse sine gives $u={\sin }^{-1}\left(0.4\right)=23.58^{\circ }$u = sin⁻¹(0.4) = 23.58°, to 2 decimal places, and the pairing gives 180° − 23.58° = 156.42°. Adding a turn gives 383.58° and 516.42°, both below 540°.

Dividing each by 3 gives x = 7.86°, 52.14°, 127.86° or 172.14°, to 2 decimal places. Check one: 3 × 52.14° = 156.42°, and sin 156.42° = 0.4000.

The usual mistakes

Searching only 0° to 360° for u. When u = 2x, the range for u runs to 720°, and stopping at 360° loses half of the solutions.

Giving values of u as the answer. For sin 2x = ½, the angles 30° and 150° are values of 2x; the answers for x are half of each.

Halving before solving. The equation is solved in u first, over the stretched range. Dividing by 2 comes last, after every value of u has been found.

Building the range in the wrong order. For u = 2x − 30°, double the range for x and then subtract 30°. Subtracting first gives a range that is wrong by 30°.