Answer
$$\eqalign{
& {\text{Concave upward}}:{\text{ }}\left( {0,\pi } \right) \cr
& {\text{Concave downward}}:{\text{ }}\left( { - \pi ,0} \right) \cr} $$
Work Step by Step
$$\eqalign{
& y = x + \frac{2}{{\sin x}},{\text{ }}\left( { - \pi ,\pi } \right) \cr
& {\text{Calculate the first and second derivatives}} \cr
& y' = \frac{d}{{dx}}\left[ {x + \frac{2}{{\sin x}}} \right] \cr
& y' = 1 - \frac{{2\cos x}}{{{{\sin }^2}x}} \cr
& y' = 1 - 2\cot x\csc x \cr
& y'' = \frac{d}{{dx}}\left[ {1 - 2\cot x\csc x} \right] \cr
& {\text{Use product rule}} \cr
& y'' = - 2\cot x\left( { - \csc x\cot } \right) - 2\csc x\left( { - {{\csc }^2}x} \right) \cr
& y'' = 2\csc x{\cot ^2}x + 2{\csc ^3}x \cr
& {\text{Set the second derivative to }}0 \cr
& 2\csc x{\cot ^2}x + 2{\csc ^3}x = 0 \cr
& {\text{Factoring}} \cr
& 2\csc x\left( {{{\cot }^2}x + 2{{\csc }^2}x} \right) = 0 \cr
& 2\csc x = 0,{\text{ }}{\cot ^2}x + 2{\csc ^2}x = 0 \cr
& 2\csc x = 0,{\text{ }}{\csc ^2}x - 1 + 2{\csc ^2}x = 0 \cr
& 2\csc x = 0,{\text{ }}\underbrace {{\text{3}}{{\csc }^2}x = 1}_{{\text{No solution}}} \cr
& \frac{2}{{\sin x}} = 0 \cr
& {\text{For the interval }}\left( { - \pi ,\pi } \right){\text{ }}\frac{2}{{\sin x}}{\text{ is not defined at }}x = 0 \cr
& {\text{We obtain the critical point }}x = 0 \cr
& {\text{Set the intervals }}\left( { - \pi ,0} \right),\left( {0,\pi } \right) \cr
& {\text{Making a table of values }}\left( {{\text{See examples on page 188 }}} \right) \cr} $$
\[\boxed{\begin{array}{*{20}{c}}
{{\text{Interval}}}&{\left( { - \pi ,0} \right)}&{\left( {0,\pi } \right)} \\
{{\text{Test Value}}}&{ - \frac{\pi }{2}}&{\frac{\pi }{2}} \\
{{\text{Sign of }}f''\left( x \right)}&{f''\left( { - \frac{\pi }{2}} \right) = - 4 > 0}&{f''\left( {\frac{\pi }{2}} \right) = 4 < 0} \\
{{\text{Conclusion}}}&{{\text{Concave downward}}}&{{\text{Concave upward}}}
\end{array}}\]
$$\eqalign{
& {\text{Concave upward}}:{\text{ }}\left( {0,\pi } \right) \cr
& {\text{Concave downward}}:{\text{ }}\left( { - \pi ,0} \right) \cr} $$
