Question

How do you solve `cosx=1+sin^2x` in the interval `0<=x<=2pi`?

Answer 1

0, `2pi`

Explanation:

Replace `sin^2 x` by `(1 - cos^2 x)`, and bring the equation to standard form:
`cos x = 1 + 1 - cos^2 x`
`cos^2 x + cos x - 2 = 0`
Solve this quadratic equation for cos x.
Since a + b + c = 0, use shortcut. There are 2 real roots:
cos x = 1 and `cos x = c/a = -2` (Rejected sin < -1)
cos x = 1 --> `x = 0 and x = 2pi`
Check
`x = 2pi` --> cos x = 1 --> sin x = 0 --> 1 = 1 + 0. OK
x = 0 --> cos x = 1 --> sin x = 0 --> 1 = 1 + 0 . OK