Question

find solutions for 2cos^2x-sinx=1

Answer 1

\[cos^2x+sin^2x=1~,~ \text{so}~~ cos^2x=1-sin^2x\]
\[2cos^2x-sinx=1\]\[2(1-sin^2x)-sinx=1\]\[2-2sin^2x-sinx=1\]\[2=2sin^2x-sinx+1\]\[0=2sin^2x+sinx-1\]\[0=(2sinx-1)(sinx+1)\]
\[2sinx-1=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~sinx+1=0\]\[2sinx=1~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~sinx=-1\]\[sinx=1/2~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x={\frac{3\pi}2} (+2\pi)\]\[x=\frac\pi6 (+2\pi) , \frac{5\pi}{6}(+2\pi)\]