Question

How do I solve for all real values of x in this equation 2 cos² x = 3 sin x ?

Answer 1

`x=pi/6+2kpi`
`x=(5pi)/6+2kpi`

Explanation:

`2cos^2x=3sinx`
`2*(1-sin^2x)=3sinx`
`2-2sin^2x=3sinx`
`2sin^2x+3sinx-2=0`

`sqrt(∆)=sqrt(25)=5`
`t_1=(-3-5)/4=-2`
`t_2=(-3+5)/4=1/2`

`sinx=1/2`

`x=pi/6+2kpi`
`x=(5pi)/6+2kpi`

k is real