Problems from Trigonometric ratios

Development:

The angle ${\displaystyle \frac{5\pi }{6}}$ is an angle in the second quadrant, that is ${\displaystyle \frac{\pi }{2}}<{\displaystyle \frac{5\pi }{6}}<\pi$, therefore we have: $$\sin (x)=\sin ({\displaystyle \frac{5\pi }{6}})=\sin (\pi -{\displaystyle \frac{5\pi }{6}})=\sin ({\displaystyle \frac{\pi }{6}})={\displaystyle \frac{1}{2}}$$

$$\cos (x)=\cos ({\displaystyle \frac{5\pi }{6}})=-\cos (\pi -{\displaystyle \frac{5\pi }{6}})=-\cos ({\displaystyle \frac{\pi }{6}})=-{\displaystyle \frac{\sqrt{3}}{2}}$$

$$\tan (x)=\tan ({\displaystyle \frac{5\pi }{6}})=-\tan (\pi -{\displaystyle \frac{5\pi }{6}})=-\tan ({\displaystyle \frac{\pi }{6}})=-{\displaystyle \frac{\sqrt{3}}{3}}$$

Solution:

$\sin (x)={\displaystyle \frac{1}{2}}$

$\cos (x)=-{\displaystyle \frac{\sqrt{3}}{2}}$

$\tan (x)=-{\displaystyle \frac{\sqrt{3}}{3}}$

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