Question

Solve the equation and write general solution: 1 + sin 2x = (sin 3x - cos 3x)²

Answer 1

Answer:

General Solution = {π/8+ 2nπ} U {3π/8+ 2nπ}  U {2nπ} U {π/2+ 2nπ}

Solution:

Given that

1 + sin 2x = (sin 3x - cos 3x)²

(sin 3x - cos 3x)² = 1 + sin 2x

sin² 3x + cos² 3x +  2 sin 3x . cos 3x = 1 + sin 2x

    1 +  2 sin 3x . cos 3x = 1 + sin 2x   ( ∵ sin² 3x + cos² 3x = 1)

Cancelling 1 on both sides, we get

2 sin 3x . cos 3x = sin 2x

sin 2(3x) = sin 2x

sin 6x - sin 2x = 0

2 cos[(6x+2x)/2]. sin[(6x-2x)/2] = 0

cos[(8x)/2]. sin[(4x)/2] = 0/2

Cos 4x . Sin 2x = 0

Cos 4x = 0 , Sin 2x = 0

For Cos 4x = 0:

Cos 4x = 0

4x = π/2    AND   4x = 3π/2

x = π/8    AND   x = 3π/8

General Solution = {π/8+ 2nπ} U {3π/8+ 2nπ}

For Sin 2x = 0:

Sin 2x = 0

2x = 0    AND   2x = π

x = 0    AND   x =π/2

General Solution = {0 + 2nπ} U {π/2+ 2nπ}

General Solution = {2nπ} U {π/2+ 2nπ}