SOLVING TRIGONOMETRIC EQUATIONS

The equations containing trigonometric functions of unknown angles are known as trigonometric equations. A solution of trigonometric equation is the value of unknown angle that satisfies the equation.

General Solution :

The solution of a trigonometric equation giving all the admissible values obtained with the help of periodicity of a trigonometric function is called the general solution of the equation.

Practice Questions

Question 1 :

Solve the following equation for which solutions lies in the interval 0° ≤ θ < 360°

sin⁴ x  =  sin² x

Solution :

sin⁴ x = sin² x

sin⁴ x - sin² x  =   0

sin² x (sin² x - 1)  =  0

Since we choose the values between 0 to 360, the solution will be {0, π/2, π, 3π/2}.

Question 2 :

Solve the following equation for which solutions lies in the interval 0° ≤ θ < 360°

2 cos² x + 1  = -3 cos x

Solution :

2 cos² x + 1  =  -3 cos x

2 cos² x + 1 + 3 cos x  =  0

Let t = cos x

2t² + 3t + 1  =  0

By factoring the quadratic equation, we get

(2t + 1)(t + 1)  =  0

For cos x  =  -1/2

For negative values of cos, we have to select the angle from 2nd and 3rd quadrants.

θ = π - a and π + a

x = π - (π/3)           x = π + (π/3)

x = 2π/3                   x = 4π/3

cos x  =  -1

x = π 

So, the solution is {π, 2π/3, 4π/3}.

Question 3 :

Solve the following equation for which solutions lies in the interval 0° ≤ θ < 360°

2 sin² x + 1  =  3sin x

Solution :

2 sin² x+1 = 3sinx

Let t = sin x

2 sin² x - 3sinx + 1 =  0

2t² - 3t + 1  =  0

By factoring the quadratic equation, we get

(2t - 1)(t - 1)  =  0

For sin x  =  1/2

x = π/6

For positive value of sin, we have to select angles from 2nd quadrant

x = π - a 

x = π - (π/6)

x = 5π/6

For sin x  = 1

x  =  π/2

So, the required solution is { π/2,  π/6,  5π/6}

Question 4 :

Solve the following equation for which solutions lies in the interval 0° ≤ θ < 360°

cos 2x  =  1 - 3 sin x

Solution :

cos 2x  =  1 - 3 sin x

 1 - 2 sin²x  =  1 - 3 sin x

2sin²x - 3 sin x + 1 - 1  =  0

2sin²x - 3 sin x  =  0

sin x (2 sin x - 3)  =  0

sin x  =  0          2 sin x - 3  = 0

x  =  sin^-1(0)          2 sin x  =  3

x  =  0, π          sin x  =  3/2

So, the solution is {0, π}.

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