How to Solve Trigonometric Functions?

Example 1: Find all the solutions of the trigonometric equation

√3 sec(θ) + 2 = 0

Solution:

Using the identity sec(θ) = 1 / cos(θ), we rewrite the equation in the form

cos(θ) = - √3 / 2

Find the reference θr angle by solving cos(θr) = √3 / 2 for θr acute.

θr = π/6

Use the reference angle θr to determine the solutions θ1 and θ2 on the interval [0 , 2π) of the given equation. The equation cos(θ) = - √3 / 2 suggests that cos(θ) is negative and that means the terminal side of angle θ solution to the given equation is either in quadrants II or III as shown below using the unit circle.

Hence the solutions:

θ1 = π - θr = π - π/6 = 5π/6

θ2 = π + θr = π + π/6 = 7π/6

Use the solutions on the interval [0 , 2π) to find all solutions by adding multiples of 2π as follows:

θ1 = 5π/6 + 2nπ , n = 0, ~+mn~ 1 , ~+mn~ 2, ...

θ2 = 7π/6 + 2nπ , n = 0, ~+mn~ 1 , ~+mn~ 2, ...

Below are shown the graphical solutions on the interval [0 , 2π)

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Example 2: Solve the trigonometric equation

2 sin(θ) = -1

Solution:

Rewrite the above equation in simple form as shown below.

sin(θ) = -1/2

Find the reference θr angle by solving sin(θ) = 1/2 for θr acute.

θr = π/6

Use the reference angle θr to determine the solutions θ1 and θ2 on the interval [0 , 2π) of the given equation. The equation sin(θ) = - 1 / 2 suggests that sin(θ) is negative and that means the terminal side of angle θ is either in quadrants III or VI as shown in the unit circle below.

Hence the solutions:

θ1 = π + θr = 7π/6

θ2 = 2π - θr = 11π/6

Use the solutions on the interval [0 , 2π) to find all solutions by adding multiples of 2π as follows:

θ1 = 7π/6 + 2nπ , n = 0, ~+mn~ 1 , ~+mn~ 2, ...

θ2 = 11π/6 + 2nπ , n = 0, ~+mn~ 1 , ~+mn~ 2, ...

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Example 3: Solve the trigonometric equation

√2 cos(3x + π/4) = - 1

Solution:

Let θ = 3x + π/4 and rewrite the equation in simple form.

√2 cos(θ) = - 1

cos(θ) = -1/√2

Find the reference θr angle by solving cos(θ) = 1/√2 for θr acute.

θr = π/4

Use the reference angle θr to determine the solutions θ1 and θ2 on the interval [0 , 2π) of the given equation. The equation cos(θ) = - 1/√2 suggests that cos(θ) is negative and that means the terminal side of angle θ is either in quadrants II or III. Hence the two solutions of the equation cos(θ) = - 1/√2 on the interval [0 , 2π) are given by

θ1 = π - θr = 3π/4

θ2 = π + θr = 5π/4

We now write the general solutions by adding multiples of 2π as follows:

θ1 = 3π/4 + 2nπ , n = 0, ~+mn~ 1 , ~+mn~ 2, ...

θ2 = 5π/4 + 2nπ , n = 0, ~+mn~ 1 , ~+mn~ 2, ...

We now substitute θ1 and θ2 by the expression 3x + π/4

3x + π/4 = 3π/4 + 2nπ

3x + π/4 = 5π/4 + 2nπ

and solve for x to obtain the solutions for x.

x = π/6 + 2nπ/3 , n = 0, ~+mn~ 1 , ~+mn~ 2, ...

x = π/3 + 2nπ/3 , n = 0, ~+mn~ 1 , ~+mn~ 2, ...

Example 4: Solve the trigonometric equation

- 2 sin^{ 2}x - cos x = - 1

Solution:

The above equation may be factored if all trigonometric functions included in that equation are the same. So using the identity sin^{ 2}x = 1 - cos^{ 2}x, we can rewrite the above equation using the same trigonometric function cos x as follows:

- 2 (1 - cos^{ 2}x) - cos x = - 1

Simplify and rewrite as

2 cos^{ 2}x - cos x - 1 = 0

Factor the left hand side

(2 cos x + 1)(cos x - 1) = 0

Hence the two equations to solve

(1) 2 cos x + 1 = 0 and (2) cos x - 1 = 0

Solve equation (1) using the reference angle as was done in the examples above.

cos x = -1/2

x_{1} = 2π/3 + 2nπ , n = 0, ~+mn~ 1 , ~+mn~ 2, ...

x_{2} = 4π/3 + 2nπ , n = 0, ~+mn~ 1 , ~+mn~ 2, ...

Solve equation (2)

cos x = 1

x_{3} = 2nπ , n = 0, ~+mn~ 1 , ~+mn~ 2, ...