Question 1105695
solve 2sin²x -5sinx-3=0
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I will solve {{{ 2sin^2(x) - 5sin(x) - 3 = 0}}} for  {{{ 0<= x <= 2pi }}}



Let u = sin(x)

Then   {{{ 2u^2 - 5u - 3 = 0 }}}
            {{{ (2u + 1)(u - 3) = 0 }}}
             {{{ u=-1/2 }}}  and  {{{ u=3 }}}

Substitute back in for u:
            {{{ sin(x) = -1/2 }}}  and {{{ sin(x) = 3 }}} 
 {{{ sin(x) = 3 }}} has no solution, discard it.

{{{sin(x) = -1/2 }}} —>  {{{  x = -pi/6 }}}  or  {{{ x = highlight((11pi)/6 ) }}}   (Q4 angle)
     There is also a Q3 angle:   {{{ x = pi + abs((-pi/6)) = highlight((7pi)/6) }}}   (Q3 angle)