Question 963442
You can't prove it because it's not an identity. 
You need to solve the equation for the values of theta for which this equation holds.
{{{2cos^2(theta)+3sin(theta)=0}}}
{{{2(1-sin^2(theta))+3sin(theta)=0}}}
{{{-2sin^2(theta)+3sin(theta)+2=0}}}
{{{2sin^2(theta)-3sin(theta)-2=0}}}
Substitute, {{{u=sin(theta)}}}
{{{2u^2-3u-2=0}}}
{{{(u-2)(2u+1)=0}}}
Two solutions:
{{{u-2=0}}}
{{{u=2}}}
{{{sin(theta)=2}}}
No solution exists.
.
.
{{{2u+1=0}}}
{{{2u=-1}}}
{{{u=-1/2}}}
{{{sin(theta)=-1/2}}}
{{{theta=210}}} and {{{theta=330}}}