Question

plz tell the answer its urgent

Answer 1

LHS of the given equation is:
2(sin6x+cos6x)−3(sin4x+cos4x)+1=2[(sin2x)3+(cos2x)3]−3(sin4x+cos4x)+1=2[(sin2x+cos2x)(sin4x+cos4x−sin2xcos2x)]−3(sin4x+cos4x)+1=2(sin4x+cos4x−sin2xcos2x)−3(sin4x+cos4x)+1=−sin4x−cos4x−2sin2xcos2x+1=−[(sin2x)2+(cos2x)2+2sin2xcos2x]+1=−(sin2x+cos2x)2+1=−12+1=−1+1=0
= RHS
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