# How do you solve sqrt3csc(9x)-7=-5?

Mar 31, 2018

$x = \frac{\pi}{27} , \frac{2 \pi}{27} , \frac{7 \pi}{27} , \frac{8 \pi}{27} , \ldots$

#### Explanation:

$\sqrt{3} \csc \left(9 x\right) - 7 = - 5$

$\sqrt{3} \csc \left(9 x\right) = 7 - 5$

$\sqrt{3} \csc \left(9 x\right) = 2$

$\csc \left(9 x\right) = \frac{2}{\sqrt{3}}$

$\csc \left(\frac{\pi}{3}\right) = \frac{2}{\sqrt{3}}$

Comparing, the fundamental value of 9x satisfying the above condition is
$9 x = \frac{\pi}{3}$

$x = \frac{\pi}{27}$
Further, the function cscx is positive in first and second quadrants

$9 x = \frac{\pi}{3} , \pi - \frac{\pi}{3} , 2 \pi + \frac{\pi}{3} , 3 \pi - \frac{\pi}{3} , \ldots .$

$9 x = \frac{\pi}{3} , \frac{2 \pi}{3} , \frac{7 \pi}{3} , \frac{8 \pi}{3} , \ldots$

$x = \frac{\pi}{27} , \frac{2 \pi}{27} , \frac{7 \pi}{27} , \frac{8 \pi}{27}$

Mar 31, 2018

$x = \frac{\pi}{27} + \frac{n 2 \pi}{9}$ , $\setminus \setminus \setminus \setminus x = \frac{2 \pi}{27} + \frac{n 2 \pi}{9}$

#### Explanation:

$\sqrt{3} \csc \left(9 x\right) - 7 = - 5$

csc(9x)=2/(sqrt(3)

Identity:

color(red)bb(csc(x)=1/sin(x)

$\frac{1}{\sin \left(9 x\right)} = \frac{2}{\sqrt{3}}$

$\sin \left(9 x\right) = \frac{\sqrt{3}}{2}$

$9 x = \arcsin \left(\sin \left(9 x\right)\right) = \arcsin \left(\frac{\sqrt{3}}{2}\right)$

$9 x = \frac{\pi}{3} + n 2 \pi$ , $\setminus \setminus \setminus \setminus 9 x = \frac{2 \pi}{3} + n 2 \pi$

$x = \frac{\pi}{27} + \frac{n 2 \pi}{9}$ , $\setminus \setminus \setminus \setminus x = \frac{2 \pi}{27} + \frac{n 2 \pi}{9}$

General solution:

For:

$n \in \mathbb{Z}$