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

##### 1 Answer

This is a quadratic equation involving trig functions.

We tackle this problem by making a substitution.

Let ...

Substitute those values into the original equation

Now we could do several things ...

1) Use integer factors. We need the factors of

2) Use the Quadratic Formula:

Let's use the first option

The factors are

**Switch back to sine**

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.

Sine corresponds to the y value in an (x,y) pair. One the unit circle y=-1 at

The only solution within the specified interval is