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

$x~=126.38° or x~=32.53°$