Solving a trigonometric cubic.

let tan(theta) = x?

Let $x = \tan(\theta)$.

First things first, it's evident that the only integer factors of 11 ($\pm$ 1 & 11) are not roots themselves. So...

The coefficient of $x^3$ isn't one, so let's divide through to make it easier to handle, albeit more clumsy-looking:



Now we're dealing with fractions. As usual, any roots will be found as a factor of the constant term. Let's try some:

To get $\frac{11}{2}$, what are the combinations you would intuitively try? Remember we are looking for only one of the roots...



That's how I'd approach it, anyway. Can't see an easier way at the moment.

What a horrible question.

Wow thanks for the help.

Change tan(theta) to x:

$[br]2 \tan ^3 \theta - 33 \tan^2 \theta - 6 \tan \theta + 11 = 0$

$[br]2x^3-33x^2-6x+11=0$

Now, since the constant is 11, there are only four possible factors, 1, -1, 11 or -11

EDIT: Are you sure you've written the question properly. Because you couldn't possibly do this in a non-calculator paper. The answers are $\tan\theta = -0.66, 0.5 or16.66$

How can I solve the following for $\theta$ without using trial and improvement to find a factor and then factorising the resulting quadratic? :
$2 \tan ^3 \theta - 33 \tan^2 \theta - 6 \tan \theta + 11 = 0$