Solve the following equations.
\cos\left(x\right) = \dfrac{-\sqrt{3}}{2}
Looking at the unit circle, we can see that the function \cos\left(x\right) has a value of \dfrac{-\sqrt{3}}{2} at \dfrac{5\pi}{6} and \dfrac{7\pi}{6}.
Since the sine function is periodic of period 2\pi, the full solution set is:
\left\{ x = \left(\dfrac{5\pi}{6}\right)+2n\pi, \ n \in \mathbb{Z} \right\} and \left\{ x = \left(\dfrac{7\pi}{6}\right)+2n\pi, \ n \in \mathbb{Z} \right\}
\cos\left(x\right) = 0
Looking at the unit circle, we can see that the function \cos\left(x\right) has a value of 0 at \dfrac{\pi}{2} and \dfrac{3\pi}{2}. Since the sine function is periodic of period 2\pi, the full solution set is:
x = \dfrac{\pi}{2}\left(2n+1\right), n \in \mathbb{Z}
x = \dfrac{\pi}{2}\left(2n+1\right), n \in \mathbb{Z}
\cos\left(x\right) = \dfrac{1}{2}
Looking at the unit circle, we can see that the function \cos\left(x\right) has a value of \dfrac{1}{2} at \dfrac{\pi}{3} and \dfrac{5\pi}{3}. Since the cosine function is periodic of period 2\pi, the full solution set is:
\left\{ x = \left(\dfrac{\pi}{3}\right)+2n\pi, n \in \mathbb{Z} \right\} and \left\{x = \left(\dfrac{5\pi}{3}\right)+2n\pi, n \in \mathbb{Z} \right\}
\cos\left(x\right) = 1
Looking at the unit circle, we can see that the function \cos\left(x\right) has a value of 1 at 0, however since the cosine function is periodic of period 2\pi, the full solution set is:
x = 0 + 2\pi n = 2\pi n, n \in \mathbb{Z}
x = 2\pi n, n \in \mathbb{Z}
x = 2\pi n, n \in \mathbb{Z}
\cos\left(x\right) = \dfrac{\sqrt{2}}{2}
Looking at the unit circle, we can see that the function \cos\left(x\right) has a value of \dfrac{\sqrt{2}}{2} at \dfrac{\pi}{4} and \dfrac{7\pi}{4}.
Since the cosine function is periodic of period 2\pi, the full solution set is:
\left\{ x = \left(\dfrac{\pi}{4}\right)+2n\pi, \ n \in \mathbb{Z} \right\} and \left\{ x = \left(\dfrac{7\pi}{4}\right)+2n\pi, \ n \in \mathbb{Z} \right\}
\cos\left(2x\right) = -1
Looking at the unit circle, we can see that the function \cos\left(x\right) has a value of -1 at \pi.
Since the cosine function is periodic of period 2\pi, the full solution set is:
2x = \pi + 2\pi n, n \in \mathbb{Z}
2x = \pi \left(2n + 1\right), n \in \mathbb{Z}
x = \pi\left(n + \dfrac{1}{2}\right), n \in \mathbb{Z}
x = \pi\left(n + \dfrac{1}{2}\right), n \in \mathbb{Z}
\cos\left(x\right) = \dfrac{-1}{2}
Looking at the unit circle, we can see that the function \cos\left(x\right) has a value of \dfrac{-1}{2} at \dfrac{2\pi}{3} and \dfrac{4\pi}{3}.
Since the cosine function is periodic of period 2\pi, the full solution set is :
\left\{ x = \left(\dfrac{2\pi}{3}\right)+2n\pi, n \in \mathbb{Z} \right\} and \left\{x = \left(\dfrac{4\pi}{3}\right)+2n\pi, n \in \mathbb{Z} \right\}