Question

What are the solutions in the interval [0,2pi)?

4sinx=-cos^2x+4

Answer 1

`x = pi/2`

Explanation:

4sin x = cos^2 x + 4.
Replace `cos^2 x` by `(1 - sin^2 x)`
`4sin x = (1 - sin^2 x) + 4`
`sin^2 x + 4sin x - 5 = 0`
Solve this quadratic equation for sin x.
Since a + b + c = 0, use shortcut. The 2 real roots are:
sin x = 1, and `sin x = c/a = -5` (rejected as < - 1)
sin x = 1 --> `x = pi/2`