Question 84904
Let {{{y=sin(x)}}}


So the equation becomes


{{{2y^2 - 6y - 3 = 0}}}


Now use the quadratic formula to solve for y

*[invoke quadratic "y", 2, -6, -3 ]


Since these values are irrational, we cannot use exact values. So we have 2 intermediate answers


{{{y=-0.436491673103709}}} or {{{y=3.43649167310371}}} 


which means


{{{sin(x)=-0.436491673103709}}} or {{{sin(x)=3.43649167310371}}}


For each case, take the arcsine of both sides

{{{arcsin(sin(x))=arcsin(-0.436491673103709)}}} or {{{arcsin(sin(x))=arcsin(3.43649167310371)}}} 


taking the arcsine of the first answer, we get

{{{x=arcsin(-0.436491673103709)=-0.45169556320258+2pi*n}}} or 


{{{x=arcsin(-0.436491673103709)=pi--0.45169556320258=3.59328821679238
}}} remember there are 2 answers when taking the arcsine

Since the domain of arcsine is *[Tex \large [-1,1]] we cannot take the arcsine of 3.436. So taking the arcsine of the second answer, we get

{{{x=arcsin(3.43649167310371)=undefined}}} 




So our answer is

{{{x=-0.45169556320258+2pi*n}}} or {{{x=3.59328821679238+2pi*n}}}