I'm not sure whether you mean to find the solution or to prove if the identity holds, but either case proves negative.
We have
#sin^2x = 4-2cos^2x=2(2-cos^2x)=2(1+color(red)(1-cos^2x))#
Note that #1-cos^color(Red)2x=sin^color(red)2x#, by the Fundamental Identity of Trigonometry.
#:. sin^2x=2(1+sin^2x)=2+2sin^2x#
#" "sin^2x=-2#
With no solutions in real numbers.
#"This is the point where we need to stop and we have clearly passed it but, let's keep going and see what happens."#
If you don't know about complex numbers nor need apply them here, then you can stop with this answer. We are doing this to find a complex solution of #x#.
The complex bit
We have #sinx=isqrt2#.
The complex definition of the sine function is
#sinz = (e^(iz)-e^(-iz))/(2i)#
Hence
#sinx=(e^(ix)-e^(-ix))/(2i)#
#isqrt2*2i=e^(ix)-e^(-ix)#
#e^(ix)+2sqrt2-1/e^(ix)=0#
Multiply both sides by #e^(ix)#:
#[e^(ix)]^2+2sqrt2e^(ix)-1=0#
This resembles a quadratic equation, doesn't it? To make it clearer, let #t=e^(ix)#.
#t^2+2tsqrt2-1=0#
with solutions #t = -sqrt2+-sqrt3#.
Back to our exponential notation,
#e^(ix)=-sqrt2+-sqrt3 => ix=ln(-sqrt2+-sqrt3)#
#x=ln(-sqrt2+-sqrt3)/i#
However, it can be shown that #1"/"i=-i#.
#:. x = -iln(-sqrt2+-sqrt3)=iln(1/(-sqrt2+-sqrt3))#
We applied #alog_b c = log_b c^a#.
Finally, rationalize the denominator:
# 1/(-sqrt2+sqrt3) = 1/(-sqrt2+sqrt3) * (-sqrt2-sqrt3)/(-sqrt2-sqrt3) = (-sqrt2-sqrt3)/(2-3)=sqrt2+sqrt3#
#1/(-sqrt2-sqrt3)=1/(-sqrt2-sqrt3) * (-sqrt2+sqrt3)/(-sqrt2+sqrt3)=(-sqrt2+sqrt3)/(2-3)=sqrt3-sqrt2#
So the set of solutions is
#color(red)(x=iln(sqrt3+-sqrt2)#
