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Question

The solution of sin8x+cos8x=1732 is

A
nπ2±π8
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B
nπ±π4
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C
nπ±π8
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D
No solution
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Solution

The correct option is A nπ2±π8

sin8(x)+cos8(x)

=(sin4x+cos4x)22sin4xcos4x

=((sin2x+cos2x)22sin2xcos2x)22sin4xcos4x

=(1sin22x2)2sin42x8

=1+sin42x4sin22xsin42x8

=1+sin42x8sin2(2x)
=1732

Therefore

Let sin2(2x)=t
Hence

t28t+1532=0

4t232t+15=0

t=32±10242408

t=32±464158

t=8±492

t=152 or t=12

Now, t=152 is not possible.

sin2(2x)ϵ[0,1]

Therefore

sin22x=12

sin(2x)=±12

2x=(2n+1)π4

Or

2x=nπ2±π8

x=nπ4±π8


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