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

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Answer 1 · 2014-10-24T22:27:44

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.