Find all the general solutions of $3 \sin 2x = -1$.

Question

Find all the general solutions of the trigonometric equation $$3 \sin 2x = -1$$

Solution. $\sin 2x = -1/3$.

I'm stuck. May somebody help please!

Answer 1

Hint:

Recall $\sin(\alpha x)$ has a period of $2\pi / \alpha$ and thus, for any integer $k$, $\sin(\alpha x) = \sin(\alpha x + 2\pi k/\alpha)$.

Noting this, take $\alpha = 2$ and take the arcsine of both sides.

Answer 2

If $$sin(A) = sin(B)$$ then the general solutions is ,

$$A = n\pi + (-1)^nB, n \in \ Z$$

Taking $A = 2x \ , \ sin(B) = -1/3 \ or \ B = - sin^{-1}(\frac{1}{3})$

$2x = n\pi + (-1)^n[- sin^{-1}(\frac{1}{3})]$

$$x = \frac{n\pi}{2} + \frac{1}{2}(-1)^{n+1}sin^{-1}\frac{1}{3} \ , \ n \in Z$$

Answer 3

I propose you the next easier (imo.) way to see things: $\;\sin x=\sin y\;$ means either $\;x=y\;$ or $\;x=\pi-y\;$, and these are the basic solutions within $\;\left[-\cfrac\pi2,\,\cfrac\pi2\right]\;$ . Then you add $\;2k\pi\;$ to each solution and we're done.

Thus, in your case:

$$\sin 2x=-\frac13\implies2x=\begin{cases}\arcsin\left(-\frac13\right)+2k\pi\approx-0.34+2k\pi (\;Rad.)\\{}\\or\\{}\\\pi-\arcsin\left(-\frac13\right)\approx 3.48+2k\pi\;(Rad.)\end{cases}\;\;\;,\;\;\;k\in\Bbb Z$$

and now just divide by $\;2\;$ and get your (infinitely many) values of $\;x\;$ .