Question

How do you find the solution to `4cos^2theta-3costheta=1` if `0<=theta<2pi`?

Answer 1

`theta = 0 or text{Arc}text{cos}(-1/4) or 2pi - text{Arc}text{cos}(-1/4)`

Explanation:

Sure you didn't mean `4cos^3 theta \ ...` ?

`4 cos ^2 theta - 3 cos theta = 1`

That's a quadratic equation in unknown `cos theta.`

`4 cos ^2 theta - 3 cos theta - 1 = 0`

It's easily factored,

`(4 cos theta + 1)(cos theta - 1) = 0`

`cos theta = -1/4 or cos theta =1`

The latter is of course `theta = 0` in the range. The former is surprisingly not one of the usual invertible angles. The best we can do is write

`theta = arccos(-1/4) = \pm text{Arc}text{cos}(-1/4) + 2 pi k quad`integer `k`

Arccos returns an angle in second quadrant for a negative argument. Putting it together our three angles in the range are

`theta = 0 or text{Arc}text{cos}(-1/4) or 2pi - text{Arc}text{cos}(-1/4)`

I don't like ruining a nice exact answer with an approximation so I'll leave the calculator stuff to others.

Check:

`4 cos ^2 0- 3 cos 0 = 4-3=1 quad sqrt`

`4 cos^2 (arccos(-1/4)) - 3 cos (arccos(-1/4)) = 4(-1/4)^2-3(-1/4)=1 quad sqrt`