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

Dec 7, 2017

$x = \pi$

#### 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 $\cos x$.

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

$1 - {\cos}^{2} x + \cos x = - 1$. This is actually a quadratic in $\cos x$!

Let $u = \cos x$, 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: $\left(u + 1\right) \left(u - 2\right) = 0$ so $u = - 1$ or $u = 2$.

Since $u = \cos x$, we now know that $\cos x = - 1$ or $\cos x = 2$. This second case is impossible, since the range of $\cos x$ is $- 1 \le y \le 1$, which does not include $2$.

$\cos x = - 1$ so $x = \arccos \left(- 1\right)$, therefore on the interval $0 \le x < 2 \pi$, $x = \pi$.

Dec 8, 2017

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

#### Explanation:

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