Question

Find all solutions of the following equation in the interval [0, 2pi).2cos^2 x = 2- sin

Answer 1

Step 1: problem

`2\cos ^2x\text{ = 2 - sinx}`

Step 2: Concept

`\text{cos}^2x=1-sin^2x`

Step 3: Method

`\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}`
`\begin{gathered} \sin x\text{ = 0 or 2sinx - 1 = 0} \\ x\text{ = }\sin ^(-1)0\text{ or 2sinx = 1} \\ \text{ x = 0 or sinx = }(1)/(2) \\ \text{ x = }\sin ^(-1)((1)/(2))\text{ = 30} \end{gathered}`
`x\text{ = 0, }\pi,\text{ }(\pi)/(6),\text{ }(5\pi)/(6)`

Final answer