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Question

The number of distinct solutions of the equation 54cos22x+cos4x+sin4x+cos6x+sin6x=2 in the interval [0,2π] is _________

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Solution

54cos22x+(cos2x)2+(sin2x)2+(cos2x)3+(sin2x)3=2
54cos22x+(cos2x+sin2x)22sin2xcos2x+(cos2x+sin2x)(sin4x+cos4xsin2xcos2x)=2
54cos22x+12sin2x.cos2x+(sin4x+cos4xsin2xcos2x)=2
54cos22x+12sin2x.cos2x+(13sin2xcos2x)=2
54cos22x5sin2x.cos2x=0
5454sin22x5sin2xcos2x=0
5454sin22x54sin22x=0
sin22x=12sin2x=±12
x=π8,3π8,5π8,7π8,9π8,11π8,13π8,15π8
Hence number of solutions in the given interval is 8.

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