Worked examples
The question
Question: solve 2 sin² x − sin x − 1 = 0.
Find every angle from 0° up to 360°.
2 sin2 x − sin x − 1 = 0
We want the angles from 0° to 360°: one full turn.
0° ≤ x < 360°
1 Name it
Look: sin² x and sin x.
Like x² and x in a quadratic.
We call sin x by one letter: u = sin x.
u = sin x
Then the equation is 2u² − u − 1 = 0: a quadratic.
We solve it the usual way.
2u2 − u − 1 = 0
2 Find two numbers
There is a number in front of u², but it can still factor.
The number in front of u² is 2.
So we multiply 2 by −1: that is −2.
We need two numbers that multiply to −2 and add to −1.
2 × (−1) = −2
? × ? = −2, ? + ? = −1
They multiply to a minus number, so one number is plus and one is minus.
They add to a minus number, so the bigger one is minus.
× is − → one +, one −; bigger one −
We list the pairs that make 2, smallest first.
We take the smaller from the bigger.
We need 1.
1 and 2 give 1.
Yes!
With the signs: 1 and −2.
1 × 2: 2 − 1 = 1 ✓
3 Split the middle
We write −u as u − 2u.
It is the same amount.
2u2 + u − 2u − 1 = 0
4 Group in pairs
We put the first two together, and the last two together.
(2u2 + u) + (−2u − 1) = 0
From the first pair we take out u.
From the second pair we take out −1.
Both leave the same bracket.
(2u2 + u) = u(2u + 1)
(−2u − 1) = −1(2u + 1)
Now both parts have 2u + 1 in them.
u(2u + 1) − 1(2u + 1) = 0
5 Same bracket out
2u + 1 is in both parts, so we take it out once.
What is left, u − 1, goes in the other bracket.
ab + cb = b(a + c)
So we get (2u + 1)(u − 1).
(2u + 1)(u − 1) = 0
6 One bracket is 0
Two numbers multiply to 0.
That only happens if one of them is 0.
So one bracket must be 0.
A × B = 0 → A = 0 or B = 0
We make each bracket 0.
Two small equations.
2u + 1 = 0 or u − 1 = 0
In 2u + 1 = 0, we take away 1 on both sides.
In u − 1 = 0, we add 1 on both sides.
2u + 1 − 1 = 0 − 1 or u − 1 + 1 = 0 + 1
Now we have 2u = −1 or u = 1.
2u = −1 or u = 1
To remove times 2, we divide both sides by 2.
So u = −1/2 or u = 1.
u = −12 or u = 1
7 Put sin x back
But u is sin x.
So sin x = −1/2 or sin x = 1.
sin x = −12 or 1
Careful!
These are not the answers yet!
They are values of sin x.
We still need the angles.
Careful: not x = −12 or 1 ✗
8 Find every angle
From the unit circle: sin 30° = 1/2.
So the reference angle is 30°.
The minus only changes which quadrants.
sin 30° = 12
Sin is the up value.
The points with that height are on a flat line.
Sin is negative below the x-axis: quadrants III and IV.
180° + 30° and 360° − 30°: x = 210° or 330°.
sin x = −12: x = 210° or 330°
The up value is 1 at one point only.
x = 90°.
sin x = 1: x = 90°
On the circle: 3 points.
All together, in order: x = 90°, 210°, 330°.
x = 90°, 210°, 330°
9 Check
Let's check every answer in 2 sin² x − sin x − 1 = 0.
sin x is 1 at 90°.
Each one must give 0.
x = 90°: 2(1)² − 1 − 1 = 2 − 1 − 1 = 0
x = 210°: 2(−12)² − (−12) − 1 = 12 + 12 − 1 = 0
x = 330°: 2(−12)² − (−12) − 1 = 12 + 12 − 1 = 0
Every answer works.
So x = 90°, 210°, 330°.
Let's look back.
First: name it.
Then: find two numbers.
Then: split the middle.
Then: group in pairs.
Then: same bracket out.
Then: one bracket is 0.
Then: put sin x back.
Then: find every angle.
Last: check.
Answer x = 90°, 210°, 330°
▶ Watch it solved
The question
Question: solve 2 cos² x + 3 cos x + 1 = 0.
Find every angle from 0° up to 360°.
2 cos2 x + 3 cos x + 1 = 0
We want the angles from 0° to 360°: one full turn.
0° ≤ x < 360°
1 Name it
Look: cos² x and cos x.
Like x² and x in a quadratic.
We call cos x by one letter: u = cos x.
u = cos x
Then the equation is 2u² + 3u + 1 = 0: a quadratic.
We solve it the usual way.
2u2 + 3u + 1 = 0
2 Find two numbers
There is a number in front of u², but it can still factor.
The number in front of u² is 2.
So we multiply 2 by 1: that is 2.
We need two numbers that multiply to 2 and add to 3.
2 × 1 = 2
? × ? = 2, ? + ? = 3
They multiply to a plus number, so the two numbers have the same sign.
They add to a plus number, so both are plus.
× is + → same signs: both +
We list the pairs that make 2, smallest first.
We add each pair.
We need 3.
1 and 2 give 3.
Yes!
1 × 2: 1 + 2 = 3 ✓
3 Split the middle
We write 3u as u + 2u.
It is the same amount.
2u2 + u + 2u + 1 = 0
4 Group in pairs
We put the first two together, and the last two together.
(2u2 + u) + (2u + 1) = 0
From the first pair we take out u.
From the second pair we take out 1.
Both leave the same bracket.
(2u2 + u) = u(2u + 1)
(2u + 1) = 1(2u + 1)
Now both parts have 2u + 1 in them.
u(2u + 1) + 1(2u + 1) = 0
5 Same bracket out
2u + 1 is in both parts, so we take it out once.
What is left, u + 1, goes in the other bracket.
ab + cb = b(a + c)
So we get (2u + 1)(u + 1).
(2u + 1)(u + 1) = 0
6 One bracket is 0
Two numbers multiply to 0.
That only happens if one of them is 0.
So one bracket must be 0.
A × B = 0 → A = 0 or B = 0
We make each bracket 0.
Two small equations.
2u + 1 = 0 or u + 1 = 0
In 2u + 1 = 0, we take away 1 on both sides.
In u + 1 = 0, we take away 1 on both sides.
2u + 1 − 1 = 0 − 1 or u + 1 − 1 = 0 − 1
Now we have 2u = −1 or u = −1.
2u = −1 or u = −1
To remove times 2, we divide both sides by 2.
So u = −1/2 or u = −1.
u = −12 or u = −1
7 Put cos x back
But u is cos x.
So cos x = −1 or cos x = −1/2.
cos x = −1 or −12
Careful!
These are not the answers yet!
They are values of cos x.
We still need the angles.
Careful: not x = −1 or −12 ✗
8 Find every angle
From the unit circle: cos 60° = 1/2.
So the reference angle is 60°.
The minus only changes which quadrants.
cos 60° = 12
Cos is the across value.
The points with that across value are on an up-and-down line.
The across value is −1 at one point only.
x = 180°.
cos x = −1: x = 180°
Cos is negative left of the y-axis: quadrants II and III.
180° − 60° and 180° + 60°: x = 120° or 240°.
cos x = −12: x = 120° or 240°
On the circle: 3 points.
All together, in order: x = 120°, 180°, 240°.
x = 120°, 180°, 240°
9 Check
Let's check every answer in 2 cos² x + 3 cos x + 1 = 0.
cos x is −1/2 at 120°.
Each one must give 0.
x = 120°: 2(−12)² + 3(−12) + 1 = 12 − 32 + 1 = 0
x = 180°: 2(−1)² + 3(−1) + 1 = 2 − 3 + 1 = 0
x = 240°: 2(−12)² + 3(−12) + 1 = 12 − 32 + 1 = 0
Every answer works.
So x = 120°, 180°, 240°.
Let's look back.
First: name it.
Then: find two numbers.
Then: split the middle.
Then: group in pairs.
Then: same bracket out.
Then: one bracket is 0.
Then: put cos x back.
Then: find every angle.
Last: check.
Answer x = 120°, 180°, 240°
▶ Watch it solved
The question
Question: solve 2 sin² x = 1 − sin x.
Find every angle from 0° up to 360°.
2 sin2 x = 1 − sin x
We want the angles from 0° to 360°: one full turn.
0° ≤ x < 360°
1 Get 0 on one side
Everything goes on the left, so the right side is 0.
We take away 1 − sin x on both sides, so they stay equal.
2 sin2 x + sin x − 1 = 0
2 Name it
Look: sin² x and sin x.
Like x² and x in a quadratic.
We call sin x by one letter: u = sin x.
u = sin x
Then the equation is 2u² + u − 1 = 0: a quadratic.
We solve it the usual way.
2u2 + u − 1 = 0
3 Find two numbers
There is a number in front of u², but it can still factor.
The number in front of u² is 2.
So we multiply 2 by −1: that is −2.
We need two numbers that multiply to −2 and add to 1.
2 × (−1) = −2
? × ? = −2, ? + ? = 1
They multiply to a minus number, so one number is plus and one is minus.
They add to a plus number, so the bigger one is plus.
× is − → one +, one −; bigger one +
We list the pairs that make 2, smallest first.
We take the smaller from the bigger.
We need 1.
1 and 2 give 1.
Yes!
With the signs: −1 and 2.
1 × 2: 2 − 1 = 1 ✓
4 Split the middle
We write u as −u + 2u.
It is the same amount.
2u2 − u + 2u − 1 = 0
5 Group in pairs
We put the first two together, and the last two together.
(2u2 − u) + (2u − 1) = 0
From the first pair we take out u.
From the second pair we take out 1.
Both leave the same bracket.
(2u2 − u) = u(2u − 1)
(2u − 1) = 1(2u − 1)
Now both parts have 2u − 1 in them.
u(2u − 1) + 1(2u − 1) = 0
6 Same bracket out
2u − 1 is in both parts, so we take it out once.
What is left, u + 1, goes in the other bracket.
ab + cb = b(a + c)
So we get (2u − 1)(u + 1).
(2u − 1)(u + 1) = 0
7 One bracket is 0
Two numbers multiply to 0.
That only happens if one of them is 0.
So one bracket must be 0.
A × B = 0 → A = 0 or B = 0
We make each bracket 0.
Two small equations.
2u − 1 = 0 or u + 1 = 0
In 2u − 1 = 0, we add 1 on both sides.
In u + 1 = 0, we take away 1 on both sides.
2u − 1 + 1 = 0 + 1 or u + 1 − 1 = 0 − 1
Now we have 2u = 1 or u = −1.
2u = 1 or u = −1
To remove times 2, we divide both sides by 2.
So u = 1/2 or u = −1.
u = 12 or u = −1
8 Put sin x back
But u is sin x.
So sin x = −1 or sin x = 1/2.
sin x = −1 or 12
Careful!
These are not the answers yet!
They are values of sin x.
We still need the angles.
Careful: not x = −1 or 12 ✗
9 Find every angle
From the unit circle: sin 30° = 1/2.
So the reference angle is 30°.
The minus only changes which quadrants.
sin 30° = 12
Sin is the up value.
The points with that height are on a flat line.
The up value is −1 at one point only.
x = 270°.
sin x = −1: x = 270°
Sin is positive above the x-axis: quadrants I and II.
30° and 180° − 30°: x = 30° or 150°.
sin x = 12: x = 30° or 150°
On the circle: 3 points.
All together, in order: x = 30°, 150°, 270°.
x = 30°, 150°, 270°
10 Check
Let's check every answer in 2 sin² x + sin x − 1 = 0.
sin x is 1/2 at 30°.
Each one must give 0.
x = 30°: 2(12)² + 12 − 1 = 12 + 12 − 1 = 0
x = 150°: 2(12)² + 12 − 1 = 12 + 12 − 1 = 0
x = 270°: 2(−1)² + (−1) − 1 = 2 − 1 − 1 = 0
Every answer works.
So x = 30°, 150°, 270°.
Let's look back.
First: get 0 on one side.
Then: name it.
Then: find two numbers.
Then: split the middle.
Then: group in pairs.
Then: same bracket out.
Then: one bracket is 0.
Then: put sin x back.
Then: find every angle.
Last: check.
Answer x = 30°, 150°, 270°
▶ Watch it solved
