Question

Determine all solutions to the equation 2cos2 x = 1 − sin x on the interval [0, 2π).

Answer 1

Given:

Expression as

$2cos^2x=1-sinx$

To find:

Determine all solutions to the equation.

Explanation:

If

$Sin\theta=Sin\alpha$

then

$\theta=n\pi+(-1)^n\alpha$

Solution:

First start as:

$\begin{gathered} 2Cos^2x-1+sinx=0 \\ 1+sinx-2sin^2x=0 \\ sinx=\frac{-1}{2},sinx=1 \end{gathered}$

So, solution will be

$x=\frac{\pi}{2},\frac{7\pi}{6},\frac{11\pi}{6}$

Hence, this is the solutions.