Finding solutions to $3\tan^2\theta-\tan\theta-14=0$ within $0\leq\theta\leq360^\circ$

Question

So the equation is $$3\tan^2\theta-\tan\theta-14=0$$ I factor to get $\tan\theta=\frac{7}{3}$, $\tan\theta=-2$. How do I find all possible solutions within $0\leq\theta\leq360^\circ$?

Using a calculator, I simply use inverse trig functions and I can only get two of the four solutions, one of which is not even correct. $\tan\theta=-2$ does not even give me an angle that satisfies the equation.

I use a Texas Instruments TI-30 II S calculator. How do I compute for all four possible angles correctly?

Answer 1

Substitute $\tan \theta =x$ then solve for $x$ then replace $x$ with $\tan \theta$ then solve for $\theta$

$3x^2-x-14=0$

$x=-2,\frac{7}{3}$

$\tan \theta = -2$

$\tan \theta = \frac{7}{3}$

$\theta =-63.43,66.8$

Add 360 to -63.43 to get to the point after a full revolution and 180 because tan is negative in second

Add 180 to 66.8 because tan is positive in first and third quadrant

$\theta = 296.66,246.8,66.8,116.56$

Answer 2

For $0\le\theta\le360$, there are only $2$ solutions for $\tan\theta =a$ that $a$ is a real number, except $a=0$, it has $3$ solutions $\theta=0^\circ,180^\circ,360^\circ$. Therefore, you have found that $\tan\theta=\dfrac{7}{3}$ or $-2$, so there are $4$ solutions. Then, you can find the solutions as Harshit Gupta