\(\sin^2x=\dfrac{1-\cos2x}2\)
\(\cos^2x=\dfrac{1+\cos2x}2\)
Using these identities, you can rewrite the equation as
\(\left(\dfrac{1-\cos2x}2\right)^3+\left(\dfrac{1+\cos2x}2\right)^3+\dfrac34\sin^22x=1\)
\(\dfrac{1-3\cos2x+3\cos^22x-\cos^32x}8+\dfrac{1+3\cos2x+3\cos^22x+\cos^32x}8+\dfrac34\sin^22x=1\)
\(\dfrac{1+3\cos^22x}4+\dfrac34\sin^22x=1\)
\(1+3\cos^22x+3\sin^22x=4\)
\(3\cos^22x+3\sin^22x=3\)
\(\cos^22x+\sin^22x=1\)
which is true for all \(x\) (infinitely many solutions).
