Solve for angles, A: $2\sin^2A+3\sin A-2=0$.

Question

Given:

$$y=2\sin^2A+3\sin A-2$$ Find all possible angles A.

Where $0\le A \le 450$ when $y=0$.

We start with, $x=\sin A$

$$2x^2+3x-2=0\tag1$$ $$(2x-1)(x+2)=0\tag2$$ $\sin A={1\over 2}$ then $A=30^{0}$

I am stuck at this point. How can I find other angles of A?

Answer 1

There are no other angles solving $x=-2$ since $\sin A\geq -1$.

Solving $x=1/2$ you have $A = 30+360k$ where $k=0,1$ and $A= 150$.

Answer 2

Remember $\sin(x)$ is a periodic function with period $T=2\pi$.

Answer 3

It is $$A=\frac{\pi}{6}+2k\pi$$ or $$A=\frac{5}{6}\pi+2k\pi$$ where $k$ is an integer number.