`2sin^2 (x) + 5cos(x) = 4` Use inverse functions where needed to find all solutions of the equation in the interval `0,2pi)`.
First transform `sin^2(x)` into `1-cos^2(x)` and obtain
This is a quadratic equation for cos(x), the roots are 2 and 1/2. cos(x) never =2.
cos(x)=1/2 on `(0, 2pi)` at `x_1=pi/3` and `x_2=(5pi)/3.`
The answer: `pi/3` and `(5pi)/3.`