Question

what the answer for solving cos^2x=3/4

Answer 1

use a calculator ma fren,use it and be amazed

Answer 2

\[\cos^2x=\frac{3}{4}\]\[\cos x=\pm \sqrt{\frac{3}{4}}=\pm \frac{\sqrt{3}}{2}\]\[x=\cos^{-1}\left(\pm \frac{\sqrt{3}}{2}\right)\]therefore \[x=\pi/6, 5\pi/6, -\pi/6,-5\pi/6, \pm 2n\pi\]where \(n~refers~to~number~of~ revolution...\)