Question 1122762
{{{sin(2A) = 1/3}}} and 2A in Q2 gives  the following picture, where we can solve for the x value of {{{-sqrt(8)}}}:<br>


{{{drawing(300,300,-4,4, -4,4, grid(0),
                line(0,0, -2.828,1),
                line(-2.828,0, -2.828,1),
               blue(locate(-3.3,0.7,"1")),
               blue(locate(-1.7,-0.3,-sqrt(8))),
               blue(locate(-1.6, 1.0 ,"3"))
)
}}} 


Now, you can use  {{{ tan(alpha + beta) = (tan(alpha)+tan(beta))/ (1-tan(alpha)tan(beta)) }}}

with {{{alpha = beta }}}:<br>

{{{ tan(2A) = -1/(sqrt(8)) = 2tan(A) / (1-tan^2(A)) }}}


which eventually (let u = tan(A),  complete the square) resolves to  {{{ highlight( tan(A) = 3 + 2sqrt(2)) }}}   in Q1