Worked examples
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 x = 15°, 75°, 195°, 255°
▶ Watch it solved
The question
Question: solve 2 sin 2x = √3.
Find every angle from 0° up to 360°.
2 sin 2x = √3
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 2 sin u = √3: the kind we can already solve.
2 sin u = √3
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 Remove the 2
2 sin u means 2 times sin u.
We want just sin u.
To remove times 2, we divide by 2.
We do it on both sides, so they stay equal.
2 sin u2 = √32
2 divided by 2 is 1.
So just sin u is left.
So sin u = √3/2.
sin u = √32
4 Find u in the first turn
From the unit circle: sin 60° = √3/2.
So the reference angle is 60°.
sin 60° = √32
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.
60° and 180° − 60°: u = 60° or 120°.
sin u = √32: u = 60° or 120°
On the circle: 2 points.
5 Add full turns
The next turn lands on the same points again: add 360°.
60° + 360° is 420°.
u = 60°, 420°
Careful!
Do not stop after one turn!
u goes round twice, so every point gives 2 angles.
Careful: not only u = 60°, 120° ✗
The same for 120°: 480°.
u = 120°, 480°
6 Divide by 2
u is 2x.
So x is u divided by 2.
x = u ÷ 2
60° ÷ 2 is 30°.
120° ÷ 2 is 60°.
And so on.
All are from 0° to 360°.
So x = 30°, 60°, 210°, 240°.
x = 30°, 60°, 210°, 240°
7 Check
Let's check every answer.
Put it in, times 2 first, then the unit circle.
Each one must give √3.
x = 30°: 2 sin(2 × 30°) = 2 sin 60° = 2 × √32 = √3
x = 60°: 2 sin(2 × 60°) = 2 sin 120° = 2 × √32 = √3
x = 210°: 2 sin(2 × 210°) = 2 sin 420° = 2 × √32 = √3
x = 240°: 2 sin(2 × 240°) = 2 sin 480° = 2 × √32 = √3
Every answer works.
So x = 30°, 60°, 210°, 240°.
Let's look back.
First: name the angle.
Then: find how far u goes.
Then: remove the 2.
Then: find u in the first turn.
Then: add full turns.
Then: divide by 2.
Last: check.
Answer x = 30°, 60°, 210°, 240°
▶ Watch it solved
The question
Question: solve sin 2x = 1/2.
Find every angle from 0 up to 2π.
sin 2x = 12
From 0 to 2π: one full turn.
We work in degrees, 0° to 360°, and change to radians at the end.
0 ≤ x < 2π
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°.
x = 15°, 75°, 195°, 255°
6 In radians
In radians: each angle times π over 180.
15° is π/12.
75° is 5π/12.
And so on.
So x = π/12, 5π/12, 13π/12, 17π/12.
x = π12, 5π12, 13π12, 17π12
7 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 = π/12, 5π/12, 13π/12, 17π/12.
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.
Then: in radians.
Last: check.
Answer x = π12, 5π12, 13π12, 17π12
▶ Watch it solved
