sin 8x cos 8x

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.

Xem thêm: dia ly duoc hinh thanh do

sin2(2x)ϵ[0,1]

Therefore 

sin22x=12

sin(2x)=±1√2

2x=(2n+1)π4

Or 

2x=nπ2±π8

⇒x=nπ4±π8