# Solve each equation, giving all solutions (general solutions), tan^5(3x)=9tan(3x)

Solve each equation, giving all solutions (general solutions), ${\mathrm{tan}}^{5}\left(3x\right)=9\mathrm{tan}\left(3x\right)$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

nitruraviX

Let's make it simple and let $y=\mathrm{tan}\left(3x\right)$, then ${y}^{5}=9y$, so $y\left({y}^{4}-9\right)=y\left({y}^{2}-3\right)\left({y}^{2}+3\right)=0.$
So $y=0,±\sqrt{3}$ are the real roots.
Now we go back to $y=\mathrm{tan}\left(3x\right)=0,±\sqrt{3}.$
So degrees. But we know that tangent is negative in QII and QIV. So $3x=180-60=120\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}x=40$ degrees, degrees.
CHECK: x=0:
x=20: 3x=60, $\mathrm{tan}\left(60\right)=\sqrt{3}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\mathrm{tan}}^{5}\left(60\right)=9\sqrt{3}=9\mathrm{tan}\left(60\right).$
x=40: 3x=120, $\mathrm{tan}\left(120\right)=-\sqrt{3}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\mathrm{tan}}^{5}\left(120\right)=-9\sqrt{3}=9\mathrm{tan}\left(120\right).$
x=100: 3x=300, $\mathrm{tan}\left(300\right)=-\sqrt{3}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\mathrm{tan}}^{5}\left(300\right)=-9\sqrt{3}=9\mathrm{tan}\left(300\right).$
But we can add 360 to 0, 60, 120 and 300, then divide by 3 to find all possible angles, so the solution is:
$x=\left(\frac{1}{3}\right)\left(360n+0\right)=120n,\left(\frac{1}{3}\right)\left(360n+60\right)=120n+20,120n+40,120n+100$ where n is an integer.