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

1 Answer
Oct 24, 2014

This is a quadratic equation involving trig functions.

We tackle this problem by making a substitution.

sin^2x-2sinx-3=0

Let ...

u=sinx

u^2=sin^2x

Substitute those values into the original equation

u^2x-2u-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=(-b+-sqrt(b^2-4ac))/(2a)

Let's use the first option

The factors are -3 and 1

(u-3)(u+1)=0

u-3=0
u=3

u+1=0
u=-1

Switch back to sine

sin(x)=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(x)=-1

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

The only solution within the specified interval is (3pi)/2 radians.