How do you solve #8(sin x)(cos x)(cos^2 x - sin^2 x) = 1#?

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Answer 1 · 2016-06-18T15:31:30

#x=(npi)/4+(-1)^n*pi/24#

#8(sin x)(cos x)(cos^2 x - sin^2 x) = 1#

Applying identities #2sinx cosx =sin2x and (cos^2 x - sin^2 x)=cos2x# we get

#4sin2xcos2x=1#

#=>2sin4x=1#

#=>sin4x=1/2= sin(pi/6)#

#=>4x=npi+(-1)^n*pi/6#

#=>x=(npi)/4+(-1)^n*pi/24#