Answer :
sin6x + cos6x = (sin2x)3 + (cos2x)3
=(sin2x + cos2x)(sin4x+cos4x-sin2xcos2x)
= 1 (sin4x + cos4x – sin2xcos2x)
Substituting above value in given equation
⇒ 2(sin4x + cos4x – sin2xcos2x)-3(sin4 x + cos4 x)+1
⇒ 2sin4x + 2cos4x – 2sin2xcos2x – 3sin4x-3cos4x+1.
⇒ -sin4x-cos4x-2sin2xcos2x+1
⇒ -[(sin2x)2+(cos2x)2-2sin2xcos2x]+1
⇒ -[( sin2x + cos2x)2]+1
⇒ -1+1
⇒ 0.
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