Question #b6403

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Answer 1 · 2017-06-19T07:21:05

#x~=126.38° or x~=32.53°#

Since #cos2x=cos^2x-sin^2x#, you get:

#cosx=2cos^2x-2sin^2x#

Then, since #sin^2x=1-cos^2x#, you get:

#cosx=2cos^2x-2+2cos^2x#

and you would solve the equation:

#4cos^2x-cosx-2=0#

by using the quadratic formula, then

#cosx=(1+-sqrt33)/8#

Then #x=cos^-1((1+-sqrt33)/8)#

#x~=126.38° or x~=32.53°#