Step 1: problem
[tex]2\cos ^2x\text{ = 2 - sinx}[/tex]
Step 2: Concept
[tex]\text{cos}^2x=1-sin^2x[/tex]
Step 3: Method
[tex]\begin{gathered} 2\cos ^2x\text{ = 2 - sinx} \\ 2(1-sin^2x)\text{ = 2 - sinx} \\ 2-2sin^2x\text{ = 2 - sinx} \\ 2-2=2sin^2x\text{ - sinx} \\ 2\sin ^2x\text{ - sinx = 0} \\ \sin x\text{ (2sinx - 1 ) = 0} \end{gathered}[/tex][tex]\begin{gathered} \sin x\text{ = 0 or 2sinx - 1 = 0} \\ x\text{ = }\sin ^{-1}0\text{ or 2sinx = 1} \\ \text{ x = 0 or sinx = }\frac{1}{2} \\ \text{ x = }\sin ^{-1}(\frac{1}{2})\text{ = 30} \end{gathered}[/tex][tex]x\text{ = 0, }\pi,\text{ }\frac{\pi}{6},\text{ }\frac{5\pi}{6}[/tex]
Final answer
