**Find All Solutions of the Equation Express the Solutions in Radians**

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.

Trigonometric equation sin θ = 0 cos θ = 0 tan θ = 0 sin θ = sinα, where α ∈ [−π/2, π/2] cos θ = cos α, where α ∈ [0,π] tan θ = tanα, where α ∈ (−π/2, π/2) |
General solution θ = nπ; n ∈ Z θ = (2n + 1) π/2; n ∈ Z θ = nπ; n ∈ Z θ = nπ + (−1) θ = 2nπ ± α, n ∈ Z θ = nπ + α, n ∈ Z |

**Question 1 :**

Solve the following equations:

(v) sin 2θ − cos 2θ − sin θ + cos θ = 0

**Solution :**

sin 2θ − cos 2θ − sin θ + cos θ = 0

sin 2θ − sin θ + cos θ − cos 2θ = 0

Let us use the formula for sin C - sin D and cos C - cos D

sin C - sin D = 2 cos (C + D)/2 sin (C - D)/2

cos C - cos D = 2 sin (C + D)/2 sin (C - D)/2

sin 2θ − sin θ = 2 cos 3θ/2 sin θ/2 -----(1)

cos θ − cos 2θ = 2 sin 3θ/2 sin θ/2 -----(2)

(1) + (2)

= 2 cos 3θ/2 sin θ/2 + 2 sin 3θ/2 sin θ/2

= 2 sin θ/2 [cos 3θ/2 + sin 3θ/2]

2 sin θ/2 = 0 sin θ/2 = 0 θ/2 = nπ θ = 2nπ |
cos 3θ/2 + sin 3θ/2 = 0 cos 3θ/2 = - sin 3θ/2 sin 3θ/2/cos 3θ/2 = -1 tan 3θ/2 = -1 a = - π/4 θ = nπ + a 3θ/2 = nπ - π/4 3θ = 2nπ - π/2 θ = 2nπ/3 - π/6 |

Hence the solution is {2nπ, 2nπ/3 - π/6}.

(vi) sin θ + cos θ = √2

**Solution :**

sin θ + cos θ = √2

Divide by √2 on both sides

(1/√2) sin θ + (1/√2) cos θ = 1

cos (π/4) cos θ - sin (π/4) sin θ = cos 0

cos (π/4 - θ) = cos 0

θ = 2nπ + a

θ = 2nπ + π/4

θ = (8n + 1)π/4

After having gone through the stuff given above, we hope that the students would have understood, "Find All Solutions of the Equation Express the Solutions in Radians"

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