Solving Trig Equations
Most trig equations have more than one solution
1. If the variable is not restricted, an equation will have an infinite number of solutions.
When all values of are required, the solution should be represented as the following where n is any integer.
 for sin and cos ,
 for tan ,
2. If an interval is given, solve for all of the values within that interval.
Assume that we are solving for all values of
in the interval
, unless otherwise stated.
Strategies:
1.
Simply find all the angles in the given interval for which the state is true by taking the inverse.
Ex:
Answer:
2.
3.
√
Note: When you want to find all values within the interval
,
Consider angles in other quadrants that have the same reference angle.
You need to use your unit circle to find these values.
and
Given sec , csc
Ex: csc = 2
and cot , change it to the reciprocal function and solve as shown above.
 sin
, find when y is
Answer:
˚
and
If an equation involves a function squared, isolate the squared term and take √ of both sides.
(Don’t forget the + and -)
Ex:

Answer: , , ,
Ex:

Answer: ,
4.
Do simple algebra whenever possible.
Ex:

√
Answer:
,
Answer:
,
 √
√
√
5.
,
√
√
Ex: √
,
Never Divide by a function of , you may lose possible solutions. Instead, add or subtract it to bring it over to
the other side, then attempt to factor the one side so that it is a product of terms equal to zero. Then use
the zero product rule.
Ex:

or
Answer: ,
˚
and ,
6.
Factor whenever possible and use the zero product rule.
Ex:

or
Answer:
,
and
Ex:

or
7.
Answer: 0, π and
Change to all the same trig functions using substitution of the identities.
Ex:

Answer: ,
8.
When all else fails, square both sides, but the answers must be substituted into the original equations to
check them since squaring could result in solutions that do not satisfy the original equations. These are
called extraneous solutions.
Ex: sin x = cos x
 sin2x = cos2x
1 – cos2x = cos2x
2 cos2x = 1
cos2x =
√
cos x =
Solutions: ,
 eliminate
,
,
and
Answer:
and
Section 1 - Find all solutions of each equation on the interval [0,2 ) .
1. 5sin x  2  sin x
2. 5  sec x  3
3. 2  4cos x  1
4. 4tan x  7  3tan x  6
5. 9  cot x  12
6. 2  10sec x  4  9sec x
7. 3csc x  2csc x  2
8. 11  3csc x  7
9. 6tan x  2  4
10. 9  sin x  10
11. 7cot x  3  4cot x
12. 7cos x  5cos x  3
2
2
2
2
2
2
Section 2 - Solve for x. Restrict the solutions to the interval [0,2 ) .
1. sin2 x  sin x  0
2. 2sin2 x  sin x  1  0
3. 2cos2 x   cos x
4. sec2 x  3sec x  2
5. tan2 x  3tan x  1  1  2tan x
6. sin3 x  sin x
7. sec x csc x  2csc x  0
8. 4sin2 x  1  0
9. 3tan3 x  tan x  0
10. 1  2cos2 x  3cos x
11. 4csc2 x  csc x  2  3csc2 x  2csc x
12. csc2 x  7csc x  2  4csc x
13. 2cot x cos2 x  cot x
14. 2sin x(sin x  1)   sin x  1
Section 3 – Solve for x. Restrict the solutions to the interval [0,2 ) .
1.
3.
5.
 7.
cot x cos2 x  cot x
2.
2cos x  3 cos x
2tan2 x sinx  tan2 x  0
csc2 x cscx 2  0
2
4.
6.
 8.
 10.
 12.
 14.
2secx 1 secx  3
2sin2 x  3sinx 1
sin2 x cscx  2sin2 x
tan2 x  3  0
9.
4sin2 x 1
(cotx 1)( 3 cot x 1)  0
3
2
 11.
cos x  cos x  2cos x
4cot x  cot x sin2 x
 13.
sin2 x  3cos2 x
2cos2 x  cos x 1 0


 When 2 or more functions occur in the same equation,collect all terms on one side and type to separate the function
if a problem originally has more than one type of trig
 by factoring or by using appropriate identities. Note, however,

function in it we need to check all solutions using the original problem. If a solution makes a part of the problem
“undefined” exclude it from your list of solutions. In other words, some factors can yield no solution.
Example 1
cot x cos2x = 2 cot x
Example 5
sec x csc x = 2 csc x
2
Example 2
2 sin x – sin x – 1 = 0
Example 6
cos x sec x + 2 cos x = 0
Example 3
2 sin2x + 3 cos x – 3 = 0
Example 7
cos x tan2x = 3 cos x
Example 4
cos x + 1 = sin x
Up until now you have been asked for solutions of trig equations within certain ranges. At the same time, you are
aware that even the simplest trig equation can have infinitely many solutions because trig functions are periodic
functions. For example, sin  = 0 is true for when  = 0, π, 2π, 3π, … and also for all negative multiples of π as well.
Overall, one could say the equation sin  = 0 has general solution  = πn where n is an integer.
For Questions 1-8, suppose a trigonometric equation has the following solutions on the interval [0,2 ) . Based on
these solutions, write an expression (or expressions) in terms of n (where n is any integer)for all solutions from
(, ) , ie, the general solutions.
1.
4.
7.
 2
4 5
0, , , ,
,
3 3
3 3
 
3
0, , , ,
4 2
2
 3 7
, ,
2 2 4
Section 4 – Find all general solutions.
1.
2sin2 x  1
4.
7.
2cos2 x  cos x  1  0
10.
cot2 x  cot x
sec x  csc x  2csc x
Section 5 – Find all general solutions.
1.
2 cos x + 1 = 0
4.
tan x + √ = 0
7.
sin2x + sin x = 0
10.
sin2x = 3 cos2x
13.
tan 3x(tan x – 1) = 0
2.
5.
8.
3 7
,
4 4
 2 7 5
, , ,
6 3 6 3

4
0, , ,
3
3
3.
6.
 5 3
, ,
6 6 2

4
, ,
4
3
2.
tan2 x  3
3.
sec2 x  1  0
5.
8.
2cos x  3  0
6.
9.
2sin2 x  1  3sin x
tan x(sec x)  tan x
2.
5.
8.
11.
14.
2 sin x + 1 = 0
3 sec2x – 4 = 0
(3 tan2x – 1)(tan2x – 3) = 0
2 sin22x = 1
cos 2x(2 cos x + 1) = 0
2csc2 x  2  5csc x
3.
6.
9.
12.
√ csc x – 2 = 0
3 cot2x – 1 = 0
4 cos2x – 1 = 0
tan23x = 3