1. ## trig solutions

From $\displaystyle (0,\pi)$

how do i solve this?

$\displaystyle 2sin(x) + csc(x) = 0$

do i subtract then divide $\displaystyle csc(x)$?

2. $\displaystyle 2\sin(x)+\csc(x)= 0$

$\displaystyle 2\sin(x)=-\csc(x)$

$\displaystyle 2\sin(x)=-\frac{1}{\sin(x)}$

now multiply both sides by sin(x) and divide both sides by 2

3. Originally Posted by pickslides
$\displaystyle 2\sin(x)+\csc(x)= 0$

$\displaystyle 2\sin(x)=-\csc(x)$

$\displaystyle 2\sin(x)=-\frac{1}{\sin(x)}$

now multiply both sides by sin(x) and divide both sides by 2
do i square both sides in the end?

4. Hello >_<SHY_GUY>_<
Originally Posted by >_<SHY_GUY>_<
From $\displaystyle (0,\pi)$

how do i solve this?

$\displaystyle 2sin(x) + csc(x) = 0$

do i subtract then divide $\displaystyle csc(x)$?
Are you sure you have this question right? pickslides has taken you as far as this:
$\displaystyle 2\sin x = -\frac{1}{\sin x}$
and has said what to do next. Well, this gives:
$\displaystyle \sin^2x = -\frac12$
In answer to your question 'do I square both sides' - no. You should take the square root now, to get $\displaystyle \sin x$ on the LHS. But can you see what the problem is with trying to take the square root of the RHS?

As it stands, then, this equation has no real solutions. Perhaps you should check that you copied it down correctly.

$\displaystyle 2\sin(x) - \csc(x) = 0$