Question

3(sin^4A+cos^4A) = 2(sin^6+cos^6)+1

Answer 1

let 3(sin^4A+cos^4A) = 2(sin^6+cos^6)+1
 ⇒ 2 ( sin^6A + cos^6A ) - 3 ( sin^4A + cos^4A ) + 1 = 0 
then,
2 ( sin^6A + cos^6A ) - 3 ( sin^4A + cos^4A ) + 1 = 0 

2 ( sin²A + cos²A ) ( sin^4A - sin²A cos²A + cos^4A ) - 3 ( sin^4A + cos^4A ) + 1 = 0 

2 ( 1 ) ( sin^4A - sin²A cos²A + cos^4A ) - 3 ( sin^4A + cos^4A ) + 1 = 0 

2 sin^4A - 2 sin²A cos²A + 2 cos^4A - 3 sin^4A - 3 cos^4A + 1 = 0 

– sin^4A - 2 sin ² A cos ² A - cos^4A + 1 = 0 

– ( sin^4A + 2 sin ² A cos ² A + cos^4A ) + 1 = 0 

– ( sin ² A + cos ² A ) ² + 1 = 0 

– ( 1 ) ² + 1 = 0 

– 1 + 1 = 0 

0 = 0

:)Hope this ans would help...