trigonometry
I am stuck when it comes to finding the end value of a trig function. I have the following question:
$$ \tan^5x – 9\tan{x} = 0 $$
I worked the problem and got:
$$ \tan x = 0\\
\tan^4x-9 = 0\\
x = 0, \pi, \frac {\pi}{3}, \frac {2\pi}{3}, \frac {4\pi}{3}, \frac {5\pi}{3} $$
My book answer is $x = \frac {\pi k}{3}$ how do you get that? I understand that tan uses $ \pi $ and sin, cos use $ 2\pi $ but I'm not sure how they got to that answer.
Making a right-triangle picture can be helpful too, either for getting to the result quickly or checking up on the analytical result. With $\tan \alpha = x = \frac{x}{1}$, we would have $x$ for the leg opposite $\alpha$ and $1$ for the adjacent leg; the hypotenuse is then $\sqrt{x^2 + 1}$ . We wish to evaluate $\sin 2\alpha$ , which is thus
$$2 \sin \alpha \cos \alpha = 2 \cdot \frac{x}{\sqrt{x^2 + 1}} \cdot \frac{1}{\sqrt{x^2 + 1}} = \frac{2x}{x^2 + 1} .$$
so $\tan\theta = \frac{a}{b} = \cot(90-\theta)$.
By the way your computation above computes $\cot 75$, not $\tan 75$.
Best Answer
$$\tan^4x=9\Longrightarrow \tan^2x= 3\Longrightarrow \tan x=\pm\sqrt 3=\pm\frac{\frac{\sqrt 3}{2}}{\frac{1}{2}}=\pm\frac{\sin\frac{\pi}{3}}{\cos\frac{\pi}{3}}$$
Can you see it now?