Question

How do you solve `(sin x)^2 + 5 cos x +5=0`?

Answer 1

`pi + 2kpi`

Explanation:

Replace in the equation `sin^2 x` by `(1 - cos^2 x)`-->
`1 - cos^2 x + 5cos x + 5 = 0`
`-cos^2 x + 5cos x + 6 = 0.`
Solve this quadratic equation for cos x.
Since a - b + c = 0, the 2 real roots are:
`cos x = -1` and `cos x = -c/a = -6/-1 = 6` (rejected as > 1)
cos x = -1 --> `x = pi`
General answers: `x = pi + 2kpi`