Question

3=2cos^(2)x+3sinx?

Answer 1

`x=pi/6,pi/2,(5pi)/6`

Explanation:

Use this identity:

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

After making this substitution, solve the problem like a quadratic.

`3=2cos^2x+3sinx`

`3=2(1-sin^2x)+3sinx`

`3=2-2sin^2x+3sinx`

`0=2-2sin^2x+3sinx-3`

`0=-2sin^2x+3sinx-1`

`0=2sin^2x-3sinx+1`

`0=(2sinx-1)(sinx-1)`

`sinx=1/2,sinx=1`

Here's a unit circle to help remind us of where the `sin` values are:

`x=pi/6,(5pi)/6,pi/2`

These are the solutions. Hope this helps!