Given:
Expression as
[tex]2cos^2x=1-sinx[/tex]
To find:
Determine all solutions to the equation.
Explanation:
If
[tex]Sin\theta=Sin\alpha[/tex]
then
[tex]\theta=n\pi+(-1)^n\alpha[/tex]
Solution:
First start as:
[tex]\begin{gathered} 2Cos^2x-1+sinx=0 \\ 1+sinx-2sin^2x=0 \\ sinx=\frac{-1}{2},sinx=1 \end{gathered}[/tex]
So, solution will be
[tex]x=\frac{\pi}{2},\frac{7\pi}{6},\frac{11\pi}{6}[/tex]
Hence, this is the solutions.
