Question

Solve the equation and write general solution: 2 sin² x + sin² 2x = 2

Answer 1

Given that

2 sin² x + sin² 2x = 2

2 sin² x + (sin 2x)² = 2

2 sin² x + (2 sin x . cos x)² = 2

2 sin² x + 4 sin² x cos² x = 2

2 sin² x + 4 sin² x (1 - sin² x) = 2

2 sin² x + 4 sin² x  - 4 sin^4 x = 2

6 sin² x  - 4 sin^4 x = 2

- 4 sin^4 x + 6 sin² x  = 2

4 sin^4 x - 6 sin² x  = - 2

4 sin^4 x - 6 sin² x  + 2 = 0

Let y = sin² x , so we have

4 y² - 6 y  + 2 = 0

Dividing both sides by 2 , we get

2 y² - 3 y  + 1 = 0

2 y² - 2 y - y + 1 = 0

2 y (y - 1) - 1(y -1) = 0

(y-1)(2y-1) = 0

y - 1 = 0 and   2y -1 = 0

y = 1  and  y = 1/2

⇒

sin² x = 1 and sin² x = 1/2

sin x = 1 , sin x = -1 , sin x = 1/√2 , sin x = - 1/√2

For sin x = 1 :-

x =  π/2

For sin x = - 1 :-

x =  3π/2

For sin x = 1/√2 :-

x =  π/4  &   x =  3π/4

For sin x = - 1/√2 :-

x =  5π/4  &   x =  7π/4

Answer 2

Solution :

2sin²x + sin² 2x = 2

=> 2sin²x + ( sin 2x )² - 2 = 0

=> 2sin²x + ( 2sinxcosx )² - 2 = 0

=> 2sin²x + 4sin²xcos²x - 2 = 0

=> 2sin²x + 4sin²xcos²x - 2 = 0

=> 2( 1 - cos²x ) + 4sin²xcos²x - 2 = 0

=> 2 - 2cos²x + 4sin²xcos²x - 2 = 0

=> -2cos²x + 4sin²xcos²x = 0

=> 2cos²x ( -1 + 2sin²x ) = 0

Therefore ,

2cos²x = 0 or -1 + 2sin²x = 0

cos² x = 0 or sin²x = 1/2

x = (2n+1)(π/2) or x = nπ±(π/4) , n € Z

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