Question

Find all solutions in the interval [0, 2pi). give exact values and show all algebra. 2cos^2 X -3sinX=0

Answer 1

using identity \(\cos^2x+\sin^2x=1\), the equation would be\[2(1-\sin^2x)-3\sin x=2-2\sin^2x-3\sin x=0\]let \(\sin x=y\) to simplify our solution\[2-2y^2-3=0\]\[2y^2+3y-2=0\]and by factoring\[(2y-1)(y+2)=0\]and since \(y=\sin x\), the sine of an angle x must be in interval \(-1\le\sin x\le1\) so \[y=\frac{1}{2}\]or \(\sin x=\frac{1}{2}\)\[x=\sin^{-1}\left(\frac{1}{2}\right)\]\[x=\frac{\pi}{6}~and~\frac{5\pi}{6}\]that is within the interval of \([0, 2\pi)\).