$\displaystyle 4cos^23x=1$

$\displaystyle cos^23x=1/4$

$\displaystyle (cos^22\pi)+cos^2x=1/4$ can i do a square root here?

assuming I cant I went on and did

$\displaystyle (2cos^2x-1)^2+cos^2x=1/4$

[tex](4cos^4x-2cos^2x+1)+cos^2x=1/4[\math]

$\displaystyle 4cos^4x-cos^2x+1=1/4$

$\displaystyle cos^2x(4cos^2x-1)+1=1/4$

how .. do I do this.. especially without the unit circle

I looked at the answers and the values are not on the unit circle.