Question

dude can someone really help me ? 3sin^2-cosx+cos^2x=0

Answer 1

plllleeeeeasssseeeee help me :(

Answer 2

\[3\sin^2x-cosx+\cos^2x=3(1-\cos^2x)-cosx+\cos^2x=-2\cos^2x-cosx+3\]\[=>cosx=-(1\pm5)/4=-1, or, cosx=-6/4<1\](From the quadratic formula). Cosx cannot be less than -1 so:\[cosx=-1=>x=\cos^{-1} -1=360n \pm180 \]where n is an integer.