Question

What are the solutions to the equation sin^2x+cosx=-1 on the interval 0 ≤ x < 2pi?

Answer 1

`x=pi`

Explanation:

To solve an equation like this, it is helpful to have only one trig function so we can solve for it and then solve for `x`. In this case it is easiest to use `cosx`.

First, we can use the Pythagorean identity `sin^2x+cos^2x=1` rearranged as `sin^2x=1-cos^2x` and plug it into the equation.

`1-cos^2x+cosx=-1`. This is actually a quadratic in `cos x`!

Let `u=cosx`, our equation is now `1-u^2+u=-1`.

Rearranging this into quadratic form we get `u^2-u-2=0`. This can be solved by factoring: `(u+1)(u-2)=0` so `u=-1` or `u=2`.

Since `u=cosx`, we now know that `cosx=-1` or `cosx=2`. This second case is impossible, since the range of `cosx` is `-1<=y<=1`, which does not include `2`.

`cosx=-1` so `x=arccos(-1)`, therefore on the interval `0<=x<2pi`, `x=pi`.

Answer 2

`0; pi/2; (2pi)/3`

Explanation:

`sin^2 x + cos x = 1`
Replace sin^2 x by `(1 - cos^2 x)` -->
`1 - cos^2 x + cos x = 1`
`- cos^2 x + cos x = 0`
(cos x)(- cos x +1) = 0
Either one of the factors should be zero.
a. cos x = 0 --> Unit circle gives 2 solutions:
`x = pi/2`, and `x = (3pi)/2`
b. - cos x + 1 = 0 --> cos x = 1
x = 0 and `x = 2pi`.
Answers for interval `[0, 2pi)`:
`0; pi/2; (2pi)/3;`
Note: 2pi is not included in the solution set --> `0 <= x < 2pi`