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

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Answer 1 · 2018-05-09T15:51:38

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

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:

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

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

How do you find the solution to 4cos^2theta-3costheta=14cos^2theta-3costheta=1 if 0<=theta<2pi0<=theta<2pi?