SOLVING TRIGNOMETRIC EQUATIONS ON THE INTERERVAL 0 AND 2PI

The equation containing trigonometric functions of unknown angles are known as trigonometric equations. A solution of a 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. We shall take the interval [−π, π] for defining the principal solution. 

Trigonometric equation

sin θ = 0

cos θ = 0

tan θ = 0

General solution

θ = nπ, n ∈ z

θ = (2n + 1)π/2, n ∈ z

θ = nπ, n ∈ z

sin θ = sin α

α ∈ [-π/2, π/2]


θ = nπ + (-1)n α, n ∈ z

cos θ = cos α

α ∈ [0, π] 


θ = 2nπ ± α, n ∈ z

tan θ = tan α

α ∈ (-π/2, π/2) 


θ = nπ + α, n ∈ z

Solve for x ∈ [0, 2π] giving your answer as exact values.

Problem 1 :

2 sin2x + sin x = 0

Solution :

2 sin2x + sin x = 0

Let t = sin x

2t2 + t = 0

t(2t + 1) = 0

t = 0 and t = -1/2

t = 0

sin x = 0

x = sin-1(0)

x = nπ

If n = 0, 1, 2

x = 0, π, 2π

sin x = -1/2

x = sin-1(-1/2)

x = -π/6  [-π/2, π/2]

x = nπ + (-1)n α

When n = 1

x = π + (-1)1 (-π/6)

x = 7π/6

n = 2

x = 2π + (-1)2 (-π/6)

x = 11π/6

So, the solutions are x = 0, π, 7π/6, 11π/6, 2π

Problem 2 :

2 cos2x = cos x

Solution :

2 cos2x - cos x = 0

cos x (2cos x - 1) = 0

cos x = 0 and 2cos x - 1 = 0

When cos x = 0

x = cos-1(0)

x = (2n + 1)π/2

n = 0, 1, ......

x = π/2, 3π/2

When 2cos x - 1 = 0

x = cos-1(1/2)

principal value

α π/3

To get multiple angles

x = 2nπ ± α

n = 0, 1, 2, .......

n = 0

x = ± π/3

n = 1

x = 2π± π/3

x = 7π/3 and x = 5π/3

Here 7π/3 is not in the interval.

x = π/3, π/2, 3π/2, 5π/3

Problem 3 :

2 cos2x + cos x - 1 = 0

Solution :

2 cos2x + cos x - 1 = 0

Let t = cos x

2t2 + t - 1 = 0

(2t - 1) (t + 1) = 0

t = 1/2 and t = -1

When cos x = 1/2

Principal value α = π/3

x = 2nπ ± α

n = 0, 1, 2, ..........

When n = 0

x = 2(0)π ± π/3

x = π/3

When n = 1

x = 2(1)π ± π/3

x = 7π/3, 5π/3

7π/3 does not belongs to the interval o to 2π.

When cos x = -1

x = π

So, the values of x are  π/3, π, π/3

Problem 4 :

2 sin2x + 3sinx + 1 = 0

Solution :

2 sin2x + 3sinx + 1 = 0

Let t = sin x

2t2 + 3t + 1 = 0

(2t + 1)(t + 1) = 0

t = -1/2 and t = -1

When sin x = -1/2

x = nπ + (-1)n α

Principal value α = -π/6

n = 0, 1, 2, ........

n = 0

x = -π/6

n = 1

x = π + (-1)1 (-π/6)

x = 7π/6

n = 2

x = 2π + (-1)2 (-π/6)

x = 11π/6

When sin x = -1

x = 3π/2

So, the values are x = 7π/6, 3π/2, 11π/6

Problem 5 :

sin2x = 2 - cos x

Solution :

sin2x = 2 - cos x

1 - cos2x = 2 - cos x

cos2x - cos x + 2 - 1 = 0

cos2x - cos x + 1 = 0

Let t = cos x

t2 - t + 1 = 0

a = 1, b = -1 and c = 1

t = -b±b2-4ac2at = -(-1)±(-1)2-4(1)(1)2(1)t = 1±-32(1)

There is no real value. So, there is no solution.

Problem 6 :

3 tan x = cot x

Solution :

3 tan x = cot x

3 tan x = 1/tan x

3tan2x = 1

tan x = 1/√3

tan x = ±1/√3

tan x = 1/√3

θ = nπ + α

Principal value α = π/6

When n = 0, 1, 2.........

n = 0

x = π/6

n = 1

x = 7π/6

tan x = -1/√3

θ = nπ + α

Principal value α = -π/6

When n = 0, 1, 2.........

n = 0

x = -π/6

n = 1

x = 5π/6

n = 2

x = 11π/6

So, the values of x are π/6, 5π/6, 7π/6, 11π/6

Problem 7 :

sin 4x = sin 2x

Solution :

sin 4x = sin 2x

sin2(2x) = sin 2x

2sin2x cos 2x - sin 2x = 0

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

sin 2x = 0 and 2 cos 2x - 1 = 0

When sin 2x = 0, x = sin-1(0)

x = 0, π/2, 3π/2, 2π

2 cos2x = 1

cos 2x = 1/2

1-2sin2x = 1/2

2sin2x = 1 - (1/2)

2sin2x = (1/2)

sin2x = (1/4)

sin x = √(1/4)

sin x = ±1/2

sin x = 1/2 and sin x = -1/2

sin x = 1/2

x = sin-1(1/2)

x = nπ + (-1)n α

Principal value α = π/6

n = 0, 1, 2, ......

n = 0

x = π/6

n = 1

x = π + (-1)1 (π/6)

x = 5π/6

sin x = -1/2

x = sin-1(-1/2)

x = nπ + (-1)n α

Principal value α = -π/6

n = 0, 1, 2, ......

n = 0

x = -π/6

n = 1

x = π + (-1)1 (-π/6)

x = 7π/6

n = 1

x = 2π + (-1)2 (-π/6)

x = 11π/6

So, the solutions are 

x = 0, π/6, π/2, 5π/6, π, 7π/6  3π/2, 11π/6, 2π

Problem 8 :

sin x + cos x = √2

Solution :

sin x + cos x = √2

Take square on both sides.

(sin x + cos x)2 = √22

sin2x + cos2x + 2sin x cos x = 2

1 - cos2x + cos2x + sin 2x = 2

sin 2x = 1

2x = sin -1(1)

2x = π/2, 3π/2

x = π/4 and 3π/4

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