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