Question

Sin^6x+cos^6x+3/4sin²2x=1

Answer 1

`\sin^2x=\frac{1-\cos2x}2`

`\cos^2x=\frac{1+\cos2x}2`

Using these identities, you can rewrite the equation as


`\left(\frac{1-\cos2x}2\right)^3+\left(\frac{1+\cos2x}2\right)^3+\frac34\sin^22x=1`

`\frac{1-3\cos2x+3\cos^22x-\cos^32x}8+\frac{1+3\cos2x+3\cos^22x+\cos^32x}8+\frac34\sin^22x=1`

`\frac{1+3\cos^22x}4+\frac34\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).