Question

$\sin^6x + \cos^6x = \frac{1}{6}$

Answer 1

The equation has no real solutions. The solution set is $\emptyset$.

Answer 2

The equation $\sin^6x + \cos^6x = \frac{1}{6}$ is simplified using the identity $\sin^6x + \cos^6x = 1 - 3\sin^2x\cos^2x$.

The equation becomes $1 - 3\sin^2x\cos^2x = \frac{1}{6}$.

Solving for $\sin^2x\cos^2x$ gives $\sin^2x\cos^2x = \frac{5}{18}$.

Using the identity $\sin^2(2x) = 4\sin^2x\cos^2x$, we get $\sin^2(2x) = 4 \cdot \frac{5}{18} = \frac{10}{9}$.

Since the maximum possible value of $\sin^2(2x)$ for real $x$ is 1, and $\frac{10}{9} > 1$, the equation $\sin^2(2x) = \frac{10}{9}$ has no real solutions.

Thus, the original equation has no real solutions. The solution set is empty.