Find the general value of $x$ satisfying the following pair. $$\sin (x)=\dfrac {1}{2} \quad\text{and}\quad \cos (x)=-\dfrac {\sqrt {3}}{2}$$
My Attempt: $$\sin (x)=\dfrac {1}{2}$$ $$\sin (x)=\sin (\dfrac {\pi}{6})$$ $$x=n\pi + (-1)^n\dfrac {\pi}{6}$$ Similarly for $\cos (x)=-\dfrac {\sqrt {3}}{2}$, $x=2n\pi\pm(\dfrac {\pi}{6})$
How do I get the general values satisfying both of them?
On
Hint: Draw the unit circle and try to locate the point $(\cos x, \sin x)$ on it. (Use the definition of $\sin$ and $\cos$)
On
For the moment, instead of $(\cos x, \sin x)$, use $(\cos \theta, \sin \theta)$.
Why? Because the point $(x,y)=(\cos \theta, \sin \theta)$ corresponds to the angle $\theta$ on the unit circle. It's confusing sometimes to use $x$ as the variable in $\cos x$ and $\sin x$ because, on the unit circle, $x$ and $y$ are coordinates of points in the plane.
From the image, you can see that the principle value of $\theta$ is $150^\circ$ or $\dfrac{5}{6}\pi$ radians.
Then the general expression for $\theta$ would be $\theta = \dfrac{5}{6}\pi + 2n\pi$ where $n$ is an integer.
Switching back to using $x$ instead of $\theta$, you would get $x = \dfrac{5}{6}\pi + 2n\pi$ where $n$ is an integer.
Hint: You can also solve $$\tan(x)=-\frac{\sqrt{3}}{3}$$