2cos2x+2cosx−1=02cos2x+2cosx−1=0 implies x = arccos( (sqrt(13) - 1)/4)x=arccos(√13−14)
(Yes; it is ugly... but the point is how to get there...)
Remember the double angle formula
cos(2x) = 2 cos^2(x) - 1cos(2x)=2cos2(x)−1
So
2cos^2x+2cosx−1=02cos2x+2cosx−1=0
becomes
4 cos^2 -2 +2 cos x - 1 = 04cos2−2+2cosx−1=0
4 cos^2 x + 2 cos x - 3 = 04cos2x+2cosx−3=0
Using the standard formula for roots of a quadratic:
(-b +- sqrt( b^2 - 4ac))/2a−b±√b2−4ac2a
We get (with minor handling)
cos(x) = ( (-1) +- sqrt(13))/4cos(x)=(−1)±√134
x = arccos ( cos(x) )x=arccos(cos(x)) by definition
noting, however, that cos(x) must fall in the range [-1, +1]
so the root ( (-1) - sqrt(13))/4(−1)−√134 can be eliminated
Therefore
x = arccos(( (-1) + sqrt(13))/4)x=arccos((−1)+√134)
