Question

Solve the following equations.
$\frac{2}{\sqrt{3}}(tanx-cotx)=ta{n}^{2}x+co{t}^{2}x-2.$

Answer 1

Question $\frac{2}{\sqrt{3}}(tanx-cotx)=ta{n}^{2}x+co{t}^{2}x-2$

Solution $\frac{2}{\sqrt{3}}(tanx-cotx)=ta{n}^{3}x+co{t}^{2}x-2tanxcotx$

$\Rightarrow \frac{2}{\sqrt{3}}(tanx-cotx)=(tanx-cotx)=$

Let $tanx-cotx=y$

$\therefore \frac{2}{\sqrt{3}}y=y^2$

$\Rightarrow y=0,\frac{2}{\sqrt{3}}$

Now,$tanx-cotx=0$

$\Rightarrow tanx=\pm 1\Rightarrow x=\pm \frac{\pi }{4}$

Now, $tanx-cotx=\frac{2}{\sqrt{3}}$

$\Rightarrow tanx-\frac{1}{tanx}=\frac{2}{\sqrt{3}}$

$\Rightarrow \sqrt{3}+ta{n}^{2}x-2tanx-\sqrt{3}=0$

$\Rightarrow tanx=\frac{-1}{\sqrt{3}},\sqrt{3}$

$\Rightarrow tanx=\frac{-1}{\sqrt{3}},\sqrt{3}\Rightarrow x=-\frac{\pi }{6},\frac{\pi }{3}$

General solution : $n\pi +x$

$n\pi +\frac{\pi }{4},n\pi -\frac{-\pi }{4},n\pi -\frac{\pi }{6},n\pi +\frac{\pi }{3}$

Solution are : $\frac{\pi }{4},\frac{-\pi }{4},\frac{-\pi }{6},\frac{\pi }{3},\frac{3\pi }{4},$

$\frac{5\pi }{4},\frac{5\pi }{6},\frac{7\pi }{6}...$