Question

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

Answer 1

`x=pi/27, (2pi)/27, (7pi)/27, (8pi)/27,...`

Explanation:

`sqrt3csc(9x)-7=-5`

`sqrt3csc(9x)=7-5`

`sqrt3csc(9x)=2`

`csc(9x)=2/sqrt3`

`csc(pi/3)=2/sqrt3`

Comparing, the fundamental value of 9x satisfying the above condition is
`9x=pi/3`

`x=pi/27`
Further, the function cscx is positive in first and second quadrants

`9x=pi/3, pi-pi/3, 2pi+pi/3, 3pi-pi/3,....`

`9x=pi/3, (2pi)/3, (7pi)/3, (8pi)/3,...`

`x=pi/27, (2pi)/27, (7pi)/27, (8pi)/27`

Answer 2

`x=pi/27+(n2pi)/9` , `\ \ \ \x=(2pi)/27+(n2pi)/9`

Explanation:

`sqrt(3)csc(9x)-7=-5`

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

Identity:

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

`1/(sin(9x))=2/(sqrt(3))`

`sin(9x)=(sqrt(3))/2`

`9x=arcsin(sin(9x))=arcsin((sqrt(3))/2)`

`9x=pi/3+n2pi` , `\ \ \ \9x=(2pi)/3+n2pi`

`x=pi/27+(n2pi)/9` , `\ \ \ \x=(2pi)/27+(n2pi)/9`

General solution:

For:

`n in ZZ`