Digital Lesson
Solving Trigonometric
Equations
sin x =
1
2
is a trigonometric equation.
x = 6π is one of infinitely many solutions of y = sin x.
y
-19π
6
-3π
-11π
6
-7π
6
π
6
1
5π
6
π
-2π -π
13π
6
17π
6
2π 3π
25π
6
4π
1
y
=
x
2
-1
All the solutions for x can be expressed in the form of
a general solution.
x = 6π + 2kπ and x = 5 6π + 2kπ (k = 0, ±1, ± 2, ± 3,  ).
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2
Find the general solution for the equation sec  = 2.
1
From cos  = 1 , it follows that cos  = .
sec 
2
1
π
1
y
cos( 3 + 2kπ) =
All values of  for which cos  =
2
P
2
are solutions of the equation.
Two solutions are  = ± π .
1
3
x
1
All angles that are coterminal
2
with ± π are also solutions and
3
can be expressed by adding integer
Q
-π + 2kπ) = 1
cos(
multiples of 2π.
3
2
The general solution can be written as  = ± π + 2kπ .
3
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3
Example: Solve tan x = 1.
The graph of y = 1 intersects the graph of y = tan x infinitely
y
many times.
- π – 2π
4
-π–π
4
π
π + π π + 2π
π + 3π
4
4
4
4
y=1
-π
π
x
2π
3π
y = tan(x)
x = -3π x = -π
2
2
x = π x = 3π x = 5π
2
2
2
Points of intersection are at x = π and every multiple of π added or
4
π
subtracted from 4 .
General solution: x = π + kπ for k any integer.
4
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4
Example:
Solve the equation 3sin x + 2 = sin x for  π ≤ x ≤ π .
2
2
y
3sin x + 2 = sin x
3sin x  sin x + 2 = 0
2sin x + 2 = 0
Collect like terms.
sin x =  2
2
1
-π
1 4
x
y=- 2
2
x = 4π is the only solution in the interval  2π ≤ x ≤ 2π .
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5
Example: To find all solutions of cos4(2x) = 9 .
16
Take the fourth root of both sides to obtain:
cos(2x)= ± 3
y
2
From the unit circle, the
solutions for 2 are
2 = ± π + kπ,
6
k any integer.
π
π
1
π
6
x
-π
6
x=- 3
2
x=
3
2
π + k ( π ), for k any integer.
Answer:  = ± 12
2
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6
Find all solutions of the trigonometric equation:
tan2  + tan  = 0.
tan2  + tan  = 0
tan  (tan  +1) = 0
Original equation
Factor.
Therefore, tan  = 0 or tan  = -1.
The solutions for tan  = 0 are the values  = kπ,
for k any integer.
The solutions for tan  = 1 are  = - 4π + kπ,
for k any integer.
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7
The trigonometric equation 2 sin2  + 3 sin  + 1 = 0 is
quadratic in form.
2 sin2  + 3 sin  + 1 = 0 implies that
(2 sin  + 1)(sin  + 1) = 0.
Therefore, 2 sin  + 1 = 0 or sin  + 1 = 0.
It follows that sin  = - 1 or sin  = -1.
2
Solutions:
 = - π + 2kπ and  = 7π + 2kπ, from sin  = - 1
6
6
2
 = -π + 2kπ, from sin  = -1
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8
Example: Solve 8 sin  = 3 cos2  with  in the
interval [0, 2π].
Rewrite the equation in terms of only one trigonometric function.
8 sin  = 3(1 sin2  )
Use the Pythagorean Identity.
3 sin2  + 8 sin   3 = 0.
A “quadratic” equation with sin x
as the variable
(3 sin   1)(sin  + 3) = 0 Factor.
Therefore, 3 sin   1 = 0 or sin  + 3 = 0
Solutions: sin  = 1 or sin  = -3
3
 = sin1( 1 ) = 0.3398 and  = π  sin1( 1 ) = 2.8107.
3
3
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s
9
Solve: 5cos2  + cos  – 3 = 0 for 0 ≤  ≤ π.
The equation is quadratic. Let y = cos  and solve 5y2 + y  3 = 0.
y = (-1 ± 61 ) = 0.6810249 or -0.8810249
10
Therefore, cos  = 0.6810249 or –0.8810249.
Use the calculator to find values of  in 0 ≤  ≤ π.
This is the range of the inverse cosine function.
The solutions are:
 = cos 1(0.6810249 ) = 0.8216349 and
 = cos 1(0.8810249) = 2.6488206
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10
Example: Find the intersection points of the graphs of
y
y = sin  and y = cos .
π + kπ
4
The two solutions
for  between 0 and
2π are π and 5π .
4
4
5 -π
4
π + kπ
4
1
-π
4
x
1
The graphs of y = sin  and y = cos  intersect at
points where sin  = cos .
This is true only for 45-45-90 triangles.
The general solution is  = 4π + kπ, for k any integer.
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