If $\sin\theta = -\dfrac{1}{2}$ , then what will be the possible values of $\theta$ between $0$ and $2π$?

Question

I even dont understood the question: If $\sin\theta = \frac{-1}{2}$ , then what will be the possible values of $\theta$ between $0$ and $ 2π $

Answer 1

An easy way of observing how many expected solutions we should get is by sketching our function $\sin(x)$ for $0\leq x \leq 2\pi$ along with $x=-\frac{1}{2}$, as follows:

We see that there are two points at which the two functions intersect, so we should expect two solutions in the interval.

We can find the first solution by using the inverse function of $\sin(x)$:

$$\sin(x)=-\frac{1}{2}\implies x=\arcsin\left(-\frac{1}{2}\right)$$

This gives us $x=-\frac{\pi}{6}$ as our principle value. We note that $\sin(x)$ is period every $2\pi$ and therefore we get: $x=\frac{11\pi}{6}$, we note that there must also be a solution at $x=\pi+(2\pi-\frac{11\pi}{6})=\frac{7\pi}{6}$.

Answer 2

Hint: First find the principle angle (i.e. what is $\sin\theta=\frac{1}{2}$, in quadrant $1$) and recall that sine is negative in quadrants $3$ and $4$.

Answer 3

sinΘ = -(1/2) at 210 and 330 radians

Answer 4

$$\sin\theta=-\frac12=-\sin\frac\pi6=\sin\left(-\frac\pi6\right)$$

$$\implies\theta=m\pi+(-1)^m\left(-\frac\pi6\right)$$ where $m$ is any integer

When $\displaystyle m$ is even, $=2r$(say) $\displaystyle\theta=2r\pi-\frac\pi6$

We need $\displaystyle 0\le2r\pi-\frac\pi6\le2\pi\implies 0<r\le1\implies r=1$

Similarly, for odd $m$