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.

Step-by-step 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=(-1)/(2),sinx=1 \end{gathered}`

So, solution will be

`x=(\pi)/(2),(7\pi)/(6),(11\pi)/(6)`

Hence, this is the solutions.