Solve 4(sin^{6}x+cos^{6}x)=4-3sin^{2} 2x

sagnuhh 2021-01-16 Answered
Solve 4(sin6x+cos6x)=43sin22x
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Expert Answer

smallq9
Answered 2021-01-17 Author has 106 answers

Firstly, work with the left side. Rewrite as a sum of two cubes:
4(sin6x+cos6x)=4((sin2x)3+(cos2x)3)
Use the rule: x3+y3=(x+y)(x2xy+y2)
4(sinx+cos6x)=4(sin2x+cos2x)(sin4xsin2xcos2x+cos4x)
Use the Pythagorean identity: sin2x+cos2x=1
4(sin6+cosx)=4(1)(sin4xsin2xcos2x+cos4)
4(sin6+cosx)=4(sin4xsin2xcos2x+cos4)
Add and substract 3sin2xcos2x to make a perfect square trinominal:
4(sin6x+cos6x)=4(sin4xsin2xcos2x+cos4x+3sin2xcos2x3sin2xcos2x)
4(sin6x+cos6x)=4(sin4x+2sin2xcos2x+cos4x3sin2xcos2x)
4(sin6x+cos6x)=4((sin2x+cos2x)23sin2xcos2x)
Use the Pythagorean identity: sin2x+cos2x=1
4(sin6x+cos6x)=4((1)23sin2xcos2x)
4(sin6x+cos6x)=4(13sin2xcos2x)
4(sin6x+cos6x)=4(13(sin2xcos2x)2)
Use the double angle identity for sine: 2x=2sinxcosx
4(sin6x+cos6)=4(13(sin2x2)2)
4(sin6+cos6)=4(13×sin22x4)

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