Question

find all the solutions in the interval [0,2pi]: 2sin^2x+5sinx-3

Answer 1

Assuming you are finding the roots (zeros) of the function, the first step is to substitute for sin x, then solve the quadratic equation, then find the values of x that cause sin x to be that value.

Answer 2

Let u = sin x, so\[2\sin^2x+5\sin x-3=0=2u^2+5u-3=(2u-1)(u+3)\]\[\implies u=-3~~~or~~~u=\frac{1}{2}\]

Answer 3

Now substitute back the other way. Note that the first solution is no good; sin x can't be -3. That means\[\sin x = \frac{1}{2} \implies x = \frac{\pi}{6}~~~or~~~x=\frac{5\pi}{6}\]

Answer 4

The key to getting this problem done is connecting what you learned in your algebra class with what you are learning in your trig class.