Solving Trigonometric Equations
How would you solve the equation sin θ = 0?
You know that θ = 0 is one solution,
but this is not the only solution. Any one of the following values of is also a
solution.
..., -3π, -2π, -π, 0, π, 2π, 3π,...
You can write this infinite solution set as {nπ:
n is an integer}
Example 1
Solving a Trigonometric Equation
Solve the equation
![](./articles_imgs/252/algebr28.gif)
Solution
To solve the equation, you should consider that the sine is negative in
Quadrants III and IV and that
![](./articles_imgs/252/algebr29.gif)
Thus, you are seeking values of θ in the third and fourth quadrants that have a reference
angle of π/3. In the interval
[0, 2π], the two angles fitting these criteria are
![](./articles_imgs/252/algebr30.gif)
By adding integer multiples of 2π to each of these solutions, you obtain the following
general solution.
See the figure below.
![](./articles_imgs/252/algebr32.gif)
Solution points of
![](./articles_imgs/252/algebr28.gif)
Example 2
Solving a Trigonometric Equation
Solve cos 2θ = 2 - 3 sin θ, where 0
≤ θ ≤ 2π.
Solution
Using the double-angle identity cos 2θ = 1 - 2
sin2θ, you can rewrite the
equation as follows.
cos 2θ
|
= 2 - 3 sin θ |
Given equation |
1 - 2 sin2θ |
= 2 - 3 sin θ |
Trigonometric identity |
0 |
= 2sin2θ - 3sin θ
+ 1 |
Quadratic form |
0 |
= (2 sin θ)(sin
θ - 1) |
Factor. |
If 2 sin θ = 0, then sin
θ = 1/2 and θ =
π/6 or θ = 5π/6. If sin θ - 1, then
sin θ = 1 and θ
= π/2. Thus, for 0 ≤ θ ≤ 2π, there are three solutions.
![](./articles_imgs/252/algebr33.gif)
|