Question 882510
Let {{{u=sin(x)}}}
{{{2u^2+u-1=0}}}
{{{(u+1)(2u-1)=0}}}
Two solutions:
{{{u+1=0}}}
{{{u=-1}}}
{{{sin(x)=-1}}}
{{{x=(3/2)pi}}}
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{{{2u-1=0}}}
{{{2u=1}}}
{{{u=1/2}}}
{{{sin(x)=1/2}}}
{{{x=pi/6}}} and {{{x=(5/6)pi}}}