How do you solve #\sin^2 x - 2 \sin x - 3 = 0# over the interval #[0,2pi]#?
This is a quadratic equation involving trig functions.
We tackle this problem by making a substitution.
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