Question 1086811
Use a substitution,
{{{u=sin(x)}}}
So,
{{{7u^2-14u+2=-5}}}
{{{7u^2-14u-7=0}}}
{{{u^2-2u-1=0}}}
Complete the square,
{{{(u^2-2u+1)-1=1}}}
{{{(u-1)^2=2}}}
{{{u-1=0 +- sqrt(2)}}}
{{{u=1 +- sqrt(2)}}}
{{{sin(x)=1 +- sqrt(2)}}}
The positive solution yields no solutions since {{{sin(x)<=1}}}
So,
{{{sin(x)=1-sqrt(2)}}}
{{{x=sin^(-1)(1-sqrt(2))}}}
Remember there are two solutions.