How do you solve \sin^2 x - 2 \sin x - 3 = 0sin2x−2sinx−3=0 over the interval [0,2pi][0,2π]?

1 Answer
AJ Speller
Oct 24, 2014

This is a quadratic equation involving trig functions.

We tackle this problem by making a substitution.

sin^2x-2sinx-3=0sin2x−2sinx−3=0

Let ...

u=sinxu=sinx

u^2=sin^2xu2=sin2x

Substitute those values into the original equation

u^2x-2u-3=0u2x−2u−3=0

Now we could do several things ...

1) Use integer factors. We need the factors of -3−3 that add up to -2−2.
2) Use the Quadratic Formula: x=(-b+-sqrt(b^2-4ac))/(2a)x=−b±√b2−4ac2a

Let's use the first option

The factors are -3−3 and 11

(u-3)(u+1)=0(u−3)(u+1)=0

u-3=0u−3=0
u=3u=3

u+1=0u+1=0
u=-1u=−1

Switch back to sine

sin(x)=3sin(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)=-1sin(x)=−1

Sine corresponds to the y value in an (x,y) pair. One the unit circle y=-1 at (3pi)/23π2 radians.

The only solution within the specified interval is (3pi)/23π2 radians.

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