Download Keeper 19 - Solving Trig Equations

Solving Trig Equations
Keeper 19
Accelerated Pre-Calculus
HELPFUL HINTS FOR SOLVING
TRIGONOMETRIC EQUATIONS
•Try to get equations in terms of one trig function by using
identities.
•Be on the look-out for ways to substitute using identities
•Try to get trig functions of the same angle. If one term is
cos2 and another is cos for example, use the double angle
formula to express first term in terms of just  instead of 2
•Get one side equals zero and factor out any common trig
functions
•See if equation is quadratic in form and will factor. (replace
the trig function with x to see how it factors if that helps)
•If the angle you are solving for is a multiple of , don't forget
to add 2 to your answer for each multiple of  since  will
still be less than 2 when solved for.
Example
Solve the equation.
2 cos x  1.
*** This is called a general solution because it shows a solution for all possible values of x.
However most problems will state the interval in which the solution is located; usually
Between 0º, 360º or 0,2𝜋 . 0,2𝜋 will be the specified interval for the remainder of the
Lesson unless otherwise stated.
2sinx  1  0
1
sinx 
2
7 11
x
,
6 6
2cos2 x  1  0
1
2
cos x 
2
 2
2
 3 5 7
x , , ,
4 4 4 4
cos x 
2cos x  3  0
3
cos x 
2
 11
x ,
6 6
tanx  1
 5
x ,
4 4
3
sinx 
2
 2
x ,
3 3
sinx  cos x
sinx cos x

cos x cos x
sinx  sinx cos x
sinx  sinx cos x  0
sinx 1  cos x   0
tanx  1
x
sinx  0
 5
,
4 4
x  0, 
cos x  1
x
sinx  tanx
sinx
sinx 
0
cos x
1 

sinx  1 
0

cos x 

sinx 1  sec x   0
sinx  0
x  0, 
sec x  1
x0
cos2x  cos x
2cos2 x  1  cos x
2cos2 x  cos x  1  0
 2cos x  1 cos x  1  0
1
cos x 
cos x  1
2
2 4 
x
,
3 3
x0
2A2  A  1   2A  1 A  1
cos2x  cos x  1  0
2cos2 x  1  cos x  1  0
2cos2 x  cos x  0
cos x  2cos x  1  0
cos x  0
1
cos x 
2
 3
x ,
2 2
2 4 
x
,
3 3
cos2 x  3sin2 x
cos2 x 3sin2 x

2
sin x
sin2 x
cot 2 x  3
cot x   3
 5 7 11
x , , ,
6 6 6 6
2sin2x  1  0
1
sin2x 
2
7 11 19 23
2x 
,
,
,
6 6
6
6
7 11 19 23
x
,
,
,
12 12 12 12
sin3x  1  0
sin3x  1
3 7 11
3x 
,
,
2 2
2
3 7 11
x
, ,
6 6 6
cos2x  7sinx  3  0
1  2sin2 x  7sinx  3  0
2sin2 x  7sinx  4  0
 2sinx  1  sinx  4   0
1
sinx  4
2
5 7
x
,
6 6
sinx 
2A2  7A  4   2A  1  A  4 
cos 2x  2cos2 x  0
2cos2 x  1  2cos2 x  0
4cos2 x  1  0
1
4
1
cos x 
2
 2 4  5 
x , , ,
3 3 3 3
cos2 x 
sin2x sinx  cos x
2sinx cos x sinx  cos x
2sin2 cos x  cos x  0


cos x 2sin2 x  1  0
1
cos x  0 sin x 
2
 2
cos x  0 sinx 
2
 3
 3 5 7
x ,
x , , ,
2 2
4 4 4 4
2
csc 2 x  cot x  1
cot 2 x  1  cot x  1
cot 2 x  cot x
cot 2 x  cot x  0
cot x  cot x  1  0
cot x  0
cot x  1
 3
x ,
2 2
 5
x ,
4 4
Example
Solve cos   sin   sin   0
2
2
Example
When we don't have squared trig
functions, we can't use the Pythagorean
identities. If you have two terms with
different trig functions you can try
squaring both sides.
cos  sin   0
Example
3tan 2x  3 in the interval 0,2 .
Example
1
cos  1  1.2108 in [0º,360º ).
2
Example
2
2
cos
u  1 cosu in 0º, 360º .
Solve
Example
2
10sin
x 12sin x  7  0 in 0º, 360º .
Solve