Solve sin 2x = 1/2 Step by Step | Multiple-angle equations

Step by step

The question

Question: solve sin 2x = 1/2.

Find every angle from 0° up to 360°.

sin 2x = 12

We want the angles from 0° to 360°: one full turn.

0° ≤ x < 360°

1 Name the angle

The angle here is 2x, not x.

We give it a name: u = 2x.

u = 2x

Then the equation is sin u = 1/2: the kind we can already solve.

sin u = 12

2 Find how far u goes

x goes from 0° to 360°.

0° ≤ x < 360°

Times 2 on every part: u = 2x goes twice as far, from 0° up to 720°.

0° ≤ 2x < 720°

720° is 2 full turns of the circle.

So u can land on each point 2 times.

720° = 2 full turns

3 Find u in the first turn

From the unit circle: sin 30° = 1/2.

So the reference angle is 30°.

sin 30° = 12

Sin is the up value.

The points with that height are on a flat line.

Sin is positive above the x-axis: quadrants I and II.

30° and 180° − 30°: u = 30° or 150°.

sin u = 12: u = 30° or 150°

On the circle: 2 points.

4 Add full turns

The next turn lands on the same points again: add 360°.

30° + 360° is 390°.

u = 30°, 390°

Careful!

Do not stop after one turn!

u goes round twice, so every point gives 2 angles.

Careful: not only u = 30°, 150° ✗

The same for 150°: 510°.

u = 150°, 510°

5 Divide by 2

u is 2x.

So x is u divided by 2.

x = u ÷ 2

30° ÷ 2 is 15°.

150° ÷ 2 is 75°.

And so on.

All are from 0° to 360°.

So x = 15°, 75°, 195°, 255°.

x = 15°, 75°, 195°, 255°

6 Check

Let's check every answer.

Put it in, times 2 first, then the unit circle.

Each one must give 1/2.

x = 15°: sin(2 × 15°) = sin 30° = 12

x = 75°: sin(2 × 75°) = sin 150° = 12

x = 195°: sin(2 × 195°) = sin 390° = 12

x = 255°: sin(2 × 255°) = sin 510° = 12

Every answer works.

So x = 15°, 75°, 195°, 255°.

Let's look back.

First: name the angle.

Then: find how far u goes.

Then: find u in the first turn.

Then: add full turns.

Then: divide by 2.

Last: check.

Answer