Question 898845
<pre>
{{{sin^2(x)+cos^2(x)=1}}}
{{{sin^2(x)+(-1/3)^2=1}}}
{{{sin^2(x)+1/9=1}}}
{{{9sin^2(x)+1=9}}}
{{{9sin^2(x)=8}}}
{{{sin^2(x)=8/9}}}
{{{sin(x)="" +- sqrt(8/9)}}}
{{{sin(x)="" +- sqrt(8)/3}}}
{{{sin(x)="" +- sqrt(4*2)/3}}}
{{{sin(x)="" +- 2sqrt(2)/3}}}

Since &#960; < x < 3&#960;/2 is quadrant III, 
we take the negative answer

{{{sin(x)=-2sqrt(2)/3}}}

{{{tan(x)=sin(x)/cos(x)=(-2sqrt(2)/3)/(-1/3)=(2sqrt(2)/3)/(1/3)=(2sqrt(2)/3)(3/1)=(2sqrt(2)/cross(3))(cross(3)/1)=2sqrt(2)}}}

Edwin</pre>