Answer to Question #221570 in Trigonometry for heyyyyy<3

Question

sec (theta)=negative two over the square root of three

solve for theta

Answer 1

$$\sec \theta = -\frac{2}{\sqrt 3}$$

Recall:

$$\sec \theta = \frac{1}{\cos \theta}$$

Therefore,

$$\frac{1}{\cos \theta} = -\frac{2}{\sqrt 3}\\ \cos \theta = -\frac{\sqrt 3}{2}\\ \theta = \cos^{-1}\Big(-\frac{\sqrt 3}{2}\Big)\\ \theta = \frac{5 \pi}{6}$$

The cosine function is negative in the second and third quadrants. To find the second solution, subtract the reference angle from $2 \pi$ to find the solution in the third quadrant.

$$\theta = 2 \pi - \frac{5 \pi}{6}\\ \theta= \frac{7 \pi}{6}$$

The period of the $\cos\theta$  function is $2 \pi$ so values will repeat every $2 \pi$ radians in both directions.

$$\therefore \qquad \theta= \frac{5 \pi}{6}+ 2 \pi n, \frac{7 \pi}{6}+ 2 \pi n \text{ (for any integer } n)$$