The solution of sin8x+cos8x=1732 is
sin8(x)+cos8(x)
=(sin4x+cos4x)2−2sin4xcos4x
=((sin2x+cos2x)2−2sin2xcos2x)2−2sin4xcos4x
=(1−sin22x2)2−sin42x8
=1+sin42x4−sin22x−sin42x8
=1+sin42x8−sin2(2x)
=1732
Therefore
Let sin2(2x)=t
Hence
t28−t+1532=0
4t2−32t+15=0
⇒t=32±√1024−2408
⇒t=32±4√64−158
⇒t=8±√492
⇒t=152 or t=12
Now, t=152 is not possible.
sin2(2x)ϵ[0,1]
Therefore
sin22x=12
sin(2x)=±1√2
2x=(2n+1)π4
Or
2x=nπ2±π8
⇒x=nπ4±π8