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Question

Solve the equation sin10x+cos10x=2916cos42x.

A
x=nπ2±π8,nz
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B
x=nπ2±π4,nz
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C
x=nπ±π8,nz
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D
x=nπ2±π2,nz
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Solution

The correct option is B x=nπ2±π4,nz
sin10x+cos10x=2916cos42xsin8xsin2x+cos8xcos2x=2916cos42x
sin8x(1cos2x)+cos8x(1sin2x)=2916cos42x
sin8x+cos8xsin8xcos2xcos8xsin2x=2916cos42x
sin8x+cos8xsin2xcos2x(sin6x+cos6x)=2916cos42x
(cos4x+sin4x)22sin4xcos4xsin2xcos2x((sin2x+cos2x)33sin2xcos2x)=2916cos42x
1sin22x+18sin42xsin2xcos2x(13sin2xcos2x)=2916cos42x
1sin22x+18sin42xsin2xcos2x+3sin4xcos2x=2916cos42x
1sin22x+18sin42x14sin22x+3sin42x(1sin2x)=2926cos42x
154sin22x+18sin42x+3sin4x3sin6x=2916cos42x
x=nπ2±π4

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