Question

Find the solution of $\sin x=\dfrac{-\sqrt{3}}{2}$

Answer 1

Hint: First we will write that for what value of sin of the angle we get $\dfrac{-\sqrt{3}}{2}$ , and then we will use the general solution of sin to find all the possible solutions, and we can see that there will be infinitely many solutions of x for which it gives $\sin x=\dfrac{-\sqrt{3}}{2}$.

Complete step-by-step answer:
Let’s first find the value of angle for which we get $\dfrac{-\sqrt{3}}{2}$.
We know that $\sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}$ ,
Now we need to find that at which quadrant sin is negative,
We know that sin is negative in ${3}^{rd}$ and ${4}^{th}$ quadrant, so if we add $\pi$ to $\dfrac{\pi }{3}$ we get,
$\pi +\dfrac{\pi }{3}=\dfrac{4\pi }{3}$
We know that $\dfrac{4\pi }{3}$ is the required value which gives $\sin x=\dfrac{-\sqrt{3}}{2}$,
Hence, we get $\sin x=\sin \dfrac{4\pi }{3}$
Now, if we have $\sin \theta =\sin \alpha$ then the general solution is:
$\theta =n\pi +{{\left( -1 \right)}^{n}}\alpha$
Now using the above formula for $\sin x=\sin \dfrac{4\pi }{3}$ we get,
$x=n\pi +{{\left( -1 \right)}^{n}}\dfrac{4\pi }{3}$ , where n = integers.
Hence, from this we can see that we will get infinitely many solutions for x.

Note: The formula for finding the general solution of sin is very important and must be kept in mind.
In the above solution we have taken the value of $\alpha$ we have taken was $\dfrac{4\pi }{3}$, but one can also take the value of $\alpha$ as $\dfrac{-\pi }{3}$ , as it lies in the ${4}^{th}$ quadrant and gives negative value for sin. And then one can use the same formula for the general solution and replace the value of $\alpha$ with $\dfrac{-\pi }{3}$ to get the answer, which is also correct.