Question

How do you find all solutions of the equation in the interval `[0,2pi)` given `2cos^2x+3cosx=0`?

Answer 1

`x={pi/2, (3pi)/2}`

Explanation:

`2cos^2x+3cosx=0`

`cosx(2cosx+3)=0`

`cosx=0 or 2cosx+3=0`

`x=cos^-1 0 or cosx=-3/2`-->Not possible

`x=cos^-1 0`

`x={pi/2, (3pi)/2}`