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.
Question 1 :
Solve the following equation for which solutions lies in the interval 0° ≤ θ < 360°
sin^{4} x = sin^{2} x
Solution :
sin^{4} x = sin^{2} x
sin^{4} x - sin^{2} x = 0
sin^{2} x (sin^{2} x - 1) = 0
sin^{2} x = 0 x = sin^{-1}(0) x = 0, π, 2π, ...... |
sin^{2} x - 1 = 0 sin^{2} x = 1 sin x = √1 x = 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 cos^{2} x + 1 = -3 cos x
Solution :
2 cos^{2} x + 1 = -3 cos x
2 cos^{2} x + 1 + 3 cos x = 0
Let t = cos x
2t^{2} + 3t + 1 = 0
By factoring the quadratic equation, we get
(2t + 1)(t + 1) = 0
2t + 1 = 0 2t = -1 t = -1/2 cos x = -1/2 |
t + 1 = 0 t = -1 cos 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 sin^{2} x + 1 = 3sin x
Solution :
2 sin^{2} x+1 = 3sinx
Let t = sin x
2 sin^{2} x - 3sinx + 1 = 0
2t^{2} - 3t + 1 = 0
By factoring the quadratic equation, we get
(2t - 1)(t - 1) = 0
2t - 1 = 0 2t = 1 t = 1/2 sin x = -1/2 |
t - 1 = 0 t = 1 sin 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 sin^{2}x = 1 - 3 sin x
2sin^{2}x - 3 sin x + 1 - 1 = 0
2sin^{2}x - 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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