Question #221570

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


solve for theta



1
Expert's answer
2021-08-09T05:44:24-0400
secθ=23\sec \theta = -\frac{2}{\sqrt 3}

Recall:


secθ=1cosθ\sec \theta = \frac{1}{\cos \theta}

Therefore,


1cosθ=23cosθ=32θ=cos1(32)θ=5π6\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 2π2 \pi to find the solution in the third quadrant.



θ=2π5π6θ=7π6\theta = 2 \pi - \frac{5 \pi}{6}\\ \theta= \frac{7 \pi}{6}

The period of the cosθ\cos\theta  function is 2π2 \pi so values will repeat every 2π2 \pi radians in both directions.


θ=5π6+2πn,7π6+2πn (for any integer n)\therefore \qquad \theta= \frac{5 \pi}{6}+ 2 \pi n, \frac{7 \pi}{6}+ 2 \pi n \text{ (for any integer } n)


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