11/21/2016
Solving Trig Equations

Ch. 5 – Analytic Trigonometry

5.3 – Solving Trig Equations
When solving trig equations, get the trig function by itself first,
then think unit circle or use a calculator to finish!
Ex: Find all solutions of 2sinx – 1 = 0 over the interval [0,
2π).
◦ Solve for sin!
◦ 2sinx = 1
◦ sinx = ½
◦ Now we can use inverse trig functions (think unit circle!)
◦ Since sinx = ½ at both  and 5 , those are our answers!
6
6
◦ Remember: x is an angle here!
Solving Trig Equations


Ex: Find all solutions of cosx + 1 = -cosx .
◦ Solve for tan!
◦ 3tan2x = 1
◦ Solve for cos!
◦ 1 = -2cosx
◦ tan2x = 1/3
◦ - ½ = cosx
◦ Find cos-1 (- ½) to get…
x
2
3
◦ tanx =
x
x
2
x
 2 n
3
and
◦ Final answer:
2
1.

4
 2 n
3
or
3. x 
4
4
5.

4
or x  
6

6

6
 n or x  

6
 n
3
2
x

3

2
or
3. x 
3
3
x
x
x
2.

4.

3  2sin x  0
Find all solutions to
over the interval [0, 2π).

x
x
4
x
 2 n
3
cos x 
Find all solutions to
2
over the interval [0, 2π).
2.
sin x
cos x
◦ Since the period of tangent is π, if we add/subtract multiples of π, we
also get solutions for x, so…
 n signifies some integer value
Final answer:

◦ However, we want ALL SOLUTIONS of x, not just on [0, 2π)!
◦ To indicate this in our answer, we add n times the period
4
1
1
 2
3
3
2
3
◦ However, we want ALL SOLUTIONS of x, not just on [0, 2π)!
◦ Since the period of sine and cosine is 2π, if we add/subtract multiples
of 2π, we also get solutions for x!
x

◦ Now we can use inverse trig functions!
◦ Find the tan-1 of both answers to get …
◦ We also want the other solution where cos(x) = -½ , so think unit circle
4
to get …
1.
Ex: Find all solutions of 3tan2x – 1 = 0 .

4
 2 n or 
or

4
7
4
 2 n
x
4.
0%
0%
0%
0%
0%
5.
x
4
5
or
3
3

6
or
5
6
0%
0%
0%
0%
0%
1
11/21/2016
Find all solutions to
1.

x
n
4
5
x  or
4
4
x

4

3.
x

4
 2 n or

4
 2 n

6
5
 2 n
4
0%
0%
0%
0%
0%
5.

◦ We can use trig identities to simplify the left side of the equation to one
trig function!
1
sin x(1  cos x) 
8
1
sin x(sin 2 x) 
8
1
3
sin x 
8

6
or 
x

6
6
or
5
6

6

x
x
.
  n or
6
n
5
n
6
0%
0%
0%
0%
0%
Ex: Find all solutions of 2sin2x – sinx – 1 = 0 on the interval
[0, 2π).
◦ Factor first…
◦ (2sinx + 1)(sinx – 1) = 0
◦ Now we have two equations to solve…
2
x

x
x
3
4
 2 n
4.
Ex: Find all solutions of sinx – sinx cos2x = 1/8 on the interval
[0, 2π).
sin x 
x
2.
3
4
x
4.
5.
or
cos 2 x 
Find all solutions to
1.

2.
3.
tan x  2  tan x .
2sin x  1  0
1
sin x  
2

6
5
6
sin x  1  0
sin x  1
◦ Think unit circle to find all values on [0, 2π) that satisfy one of those
equations…
x
1
2
7
6
x
11
6
x

2
◦ All three of these are possible answers!

Ex: Find all solutions of 4cos2(3x) – 1 = 0 .

3


Ex: Find all solutions of csc (4x) – 2 = 0 on the interval [0, π).
◦ Solve for csc, then flip it to solve for sin…
◦ csc(4x) = 2
◦ cos2(3x) = 1/4
◦ cos(3x) = ± ½
◦ Now we have two equations to solve. However, we won’t divide by 3
until the trig function has left the equation…
◦ sin(4x) = ½
◦ Using unit circle knowledge, we know…
cos 3x 
3x 

◦ Solve for cos…
◦ 4cos2(3x) = 1
 2 n
1
2
5
3x 
 2 n
3
cos 3x  
3x 
2
 2 n
3
1
2
4
3x 
 2 n
3
Final answers:
2 n
x 
9
3
5 2 n
x

9
3
2 2 n
x

9
3
4 2 n
x

9
3
4x 
x

6

24
 2 n

n
2
4x 
5
 2 n
6
x
5  n

24 2
◦ Now find all solutions to these equations on the interval [0, π)…
x
 13
,
24 24
x
5 17
,
24 24
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Find all solutions to
1.
3
n
4
3  n
x

8
2
x
2.
3.
x

4
1.
3.
3
x
n
8
x

8
 2 n
x

3
x
2.
 n
4.
5.
sec x  2  0
Find all solutions to
over the interval [0, 2π).
3tan 2 x  3  0.
x

3
or 
0%
0%
0%
0%
0%
5.
x

3
3
 2 n

3
x
4.


3
 2 n or 
or

3
5
3
 2 n
0%
0%
0%
0%
0%
3