# 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.

Principal Solution

The smallest numerical value of unknown angle satisfying the equation in the interval [0, 2π] (or) [−π, π] is called a principal solution.

Principal value of sine function lies in the interval

[−π/2, π/2]

and hence lies in I quadrant or IV quadrant.

Principal value of cosine function is in

[0, π]

and hence in I quadrant or II quadrant.

Principal value of tangent function is in

(-π/2, π/2)

and hence in I quadrant or IV quadrant.

## 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

 sin2 x  =  0 x = sin-1(0)x = 0, π, 2π, ...... sin2 x - 1  =  0sin2 x =  1sin x = √1x = sin-1(1) (or) x = sin-1 (-1)x = π/2 and x = 3π/2

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

 2t + 1  =  02t  =  -1t  =  -1/2cos x  =  -1/2 t + 1  =  0t = -1cos x  = -1

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

 2t - 1  =  02t  =  1t  =  1/2sin x  =  -1/2 t - 1  =  0t  =  1sin x  =  1

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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