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Question

Solve the following equations.
4cosx2cos2xcos4x=1.

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Solution

4cosx2cos2xcos4x=1
4cosxcos2xcos4x=1+cos2x
4cosx2cos3xcosx=2cos2x
Say cosx=t
4t2t(4t33t)=2t2
4t8t4+6t2=2t2
8t44t24t=0
2t4t2t=0
t(t1)(2t2+2t+1)=0
t=0 or t=1
cosx=0 or cosx=1
x=(2x+1)π/2x=2,mπ
nϵ integer mϵ integer

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