Question

2 (sin6A + cos6A ) -3 (sin4A + cos4A) + 1 = 0​

Answer 1

Solution :-

→ 2(sin6a+cos6a) -3(sin4a+cos4a)+1=0

→ 2[(sin2a)3+(cos2a)3] -3[(sin2a)2+(cos2a)2]+1

→ 2[(sin2a+cos2a)3] -3sin2acos2a(sin2a+cos2a)]

→ -2sin2acos2a]+1 - 3[(sin2a+cos2a)2

→ 2[1-3sin2acos2a] -3[1-2sin2acos2a]+1

→ 2-6sin2acos2a -3+6sin2acos2a+1

→ 0