Question

Question #7ee2f

Answer 1

The solutions are `S={0, pi, 7/6pi, 11/6pi}`

Explanation:

We need

`cos^2x+sin^2x=1`

Our equation is

`2cos^2x=2+sinx`

Therefore,

`2(1-sin^2x)=2+sinx`

`2-2sin^2x=2+sinx`

`2sin^2x+sinx=0`

Factorising

`sinx(2sinx+1)=0`

so,

`sinx=0`, `=>`, `x=0` or `x=pi`

`sinx=-1/2`, `=>`, `x=7/6pi` or `x=11/6pi`

`AA x in [0, 2pi)`