If the equation is
[tex]7\sin^2x-14\sin x+2=-5[/tex]
then rewrite the equation as
[tex]7\sin^2x-14\sin x+7=0[/tex]
Divide boths sides by 7:
[tex]\sin^2x-2\sin x+1=0[/tex]
Since [tex]x^2-2x+1=(x-1)^2[/tex], we can factorize this as
[tex](\sin x-1)^2=0[/tex]
Now solve for x :
[tex]\sin x-1=0[/tex]
[tex]\sin x=1[/tex]
[tex]\implies\boxed{x=\dfrac\pi2+2n\pi}[/tex]
where n is any integer.
If you meant sin(2x) instead, I'm not sure there's a simple way to get a solution...
