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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