Question

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

Answer 1

`\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})`