1. ## Trigo Angles

Solve the equation 6x^3 + 11x^2 - 3x - 2 = 0
Hence, find the values of angle [D] between 0 and 180 degrees, which satisfy the equation 6 tan[D]^2 + 11 tan [D] - 3 = 2 cot [D]

I have solved the cubic equation by factor theorem - its answers are
-2, -1/3, 1/2

I can't seem to find the values of angle [D], since the new equation doesn't exactly tally with the cubic one.

Thank you.

2. Originally Posted by Drdj
Solve the equation 6x^3 + 11x^2 - 3x - 2 = 0
Hence, find the values of angle [D] between 0 and 180 degrees, which satisfy the equation 6 tan[D]^2 + 11 tan [D] - 3 = 2 cot [D]

I have solved the cubic equation by factor theorem - its answers are
-2, -1/3, 1/2

I can't seem to find the values of angle [D], since the new equation doesn't exactly tally with the cubic one.

Thank you.
Yes it does...

$\displaystyle 6\tan^2{D} + 11\tan{D} - 3 = 2\cot{D}$

Multiply everything by $\displaystyle \tan{D}$

$\displaystyle 6\tan^3{D} + 11\tan^2{D} - 3\tan{D} = 2$

$\displaystyle 6\tan^3{D} + 11\tan^2{D} - 3\tan{D} - 2 = 0$.

Now by letting $\displaystyle x = \tan{D}$, you have the same cubic equation...

3. ## Thanks

Thank you so much !! Never would have thought of that !!

4. Thanks Prove It, I learnt something too!!