# How do you solve \sin^2 x - 2 \sin x - 3 = 0 over the interval [0,2pi]?

Oct 24, 2014

This is a quadratic equation involving trig functions.

We tackle this problem by making a substitution.

${\sin}^{2} x - 2 \sin x - 3 = 0$

Let ...

$u = \sin x$

${u}^{2} = {\sin}^{2} x$

Substitute those values into the original equation

${u}^{2} x - 2 u - 3 = 0$

Now we could do several things ...

1) Use integer factors. We need the factors of $- 3$ that add up to $- 2$.
2) Use the Quadratic Formula: $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Let's use the first option

The factors are $- 3$ and $1$

$\left(u - 3\right) \left(u + 1\right) = 0$

$u - 3 = 0$
$u = 3$

$u + 1 = 0$
$u = - 1$

Switch back to sine

$\sin \left(x\right) = 3$

The above is an erroneous solution because sin oscillates between 1 and -1. Sine will never result in 3 so we disregard this as a solution.

$\sin \left(x\right) = - 1$

Sine corresponds to the y value in an (x,y) pair. One the unit circle y=-1 at $\frac{3 \pi}{2}$ radians.

The only solution within the specified interval is $\frac{3 \pi}{2}$ radians.