Finding ALL solutions to $2(\sin^2(x)) - 5\sin(x)-3 = 0$?

Question

What does it mean to find "ALL possible solutions?"

I know it has something to do with simplifying the equation, getting the angle (in radians) by doing the inverse.. and adding $2\pi n$?

So given the interval $[0,2\pi]$ and the equation:

$2(\sin^2(x)) - 5\sin(x)-3 = 0$

I factored the equation to get:

$ (\sin(x)-3)(1+2\sin(x)) = 0$

Then I set:

$(\sin(x)-3) = 0$ or $(1+2\sin(x)) = 0$

I simplified them to:

$\arcsin(3) = x$ or $\arcsin(-1/2) = x$

I then got $\arcsin(-1/2) = (11(\pi))/6 + 2\pi(n)$

But how do I get the solution for$\arcsin(3)$?

Thanks!

Answer 1

You factored correctly, but note that there is no real $x$ such that $\sin x = 3$.
$|\sin x| \leq 1$, for all $x$. That is, it will never be the case that the factor $(\sin x - 3) = 0$.

But there is another solution for which $\sin x = -\dfrac 12,\;$ with $x \in [0, 2\pi]$.

Hint: consider the third quadrant: $x = \pi \,+ \;?\;$

Answer 2

By changing variable, couldn't you solve this ?

$X = sin(X)$

So now, you want to solve

$$2X^{2}-5X-3=0$$ $$\Delta = 5^2-4*2*(-3) $$ $$\Delta = 49$$

Therefore you get $X_{1} = -\frac{1}{2}$ and $X_{2} = 3$

Which is $sin(X)=-\frac{1}{2}$ and $sin(X)=3$

As $-1\le sin(X) \le 1$, the second solution is not valid.

$$sin(X_{1}) = -\frac{1}{2}$$

Which leads to $X_{1}= \frac{7\pi}{6}[2\pi]$ and $X_{12}=\frac{11\pi}{6}[2\pi]$ I may have done horrible mistakes here, please feel free to correct me !