How do you solve the following equation #-6 cos( pi/3) theta = 4# in the interval [0, 2pi]?

1 Answer
KillerBunny
May 2, 2018

#\theta = \pm\frac{3}{\pi} * \arccos(-\frac{2}{3})#

Explanation:

If the equation is as you wrote, and thus #\theta# is not part of the argument of the cosine, you simply have #\cos(\pi/3) = 1/2# and thus the equation becomes

#-6 * 1/2 \theta = 4#

#-3\theta = 4#

#\theta = -4/3#

and thus no solutions in #[0,2\pi]#

However, I presume that the equation was actually

#-6cos(\theta\frac{\pi}{3})=4#

In this case, we isolate the cosine:

#cos(\theta\frac{\pi}{3})=-\frac{4}{6} = -\frac{2}{3}#

This means that

#\theta\frac{\pi}{3} = \pm\arccos(-\frac{2}{3})#

and thus

#\theta = \pm\frac{3}{\pi} * \arccos(-\frac{2}{3})#

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