Question 1045455
{{{(2(cosx)^2 + (sinx) -(1))}}}={{{0}}}
since 

{{{((cosx)^2 + (sinx)^2 )}}}={{{1}}}
{{{((1- (sinx)^2 ))}}}={{{(cosx)^2}}}

{{{(2(1-(sinx)^2) + (sinx) -(1))}}}={{{0}}}

{{{((2-2(sinx)^2) + (sinx) -(1))}}}={{{0}}}
{{{((1-2(sinx)^2) + (sinx) )}}}={{{0}}}
multiply {{{-1}}} both sides

{{{((2(sinx)^2) - (sinx)-1 )}}}={{{0}}}



let {{{u}}}={{{sinx}}}
{{{((2(u)^2) - (u) -1)}}}={{{0}}}
{{{(2*u+1) * (u -1)}}}={{{0}}}

{{{(2*u+1)}}}={{{0}}} or {{{((u) -1)}}}={{{0}}}

{{{2(u)}}}={{{-1}}} or {{{u }}}={{{1}}}
{{{u}}}={{{-1/2}}} or {{{u }}}={{{1}}}
{{{(2sinx+1)*(sinx-1)}}}={{{0}}} 
{{{(2sinx+1)}}}={{{0}}} or {{{(sinx-1)}}}={{{0}}} 
{{{(2sinx)}}}={{{-1}}} or {{{(sinx)}}}={{{1}}} 
{{{(sinx)}}}={{{-1/2}}} or {{{(sinx)}}}={{{1}}} 
{{{x}}}={{{7pi/6}}},{{{x}}}={{{11pi/6}}}
or {{{x}}}={{{pi/2}}}