Question

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

Answer 1

x=0

Explanation:

`sin^2 x=1-cos^2 x`
`1-cos^2 x + 2 cos x = 2`
let `A=cosx`
Now solve
`-A^2+2A-1=0`
`-(A-1)^2=0`
`A=cosx=1`
`x=2npi`(n=integer number)
In the interval `[0,2pi)`
the only solution is
when n=0 x=0