Solve the equation $2\sin^4 x + 7\cos^2 x = 4$ for $0^\circ \le x \le 360^\circ$.

1. Express the equation in terms of a single trigonometric function

Substitute $\cos^2 x = 1 - \sin^2 x$ into the equation:
$2\sin^4 x + 7(1 - \sin^2 x) = 4$
$2\sin^4 x - 7\sin^2 x + 7 - 4 = 0$
$2\sin^4 x - 7\sin^2 x + 3 = 0$

2. Solve the quadratic equation in $\sin^2 x$

Let $u = \sin^2 x$. The equation becomes:
$2u^2 - 7u + 3 = 0$
$(2u - 1)(u - 3) = 0$
$u = \frac{1}{2} \text{ or } u = 3$
So, $\sin^2 x = \frac{1}{2}$ or $\sin^2 x = 3$.

3. Find the possible values of $\sin x$

Since $-1 \le \sin x \le 1$, $\sin^2 x = 3$ has no real solutions (reject).
For $\sin^2 x = \frac{1}{2}$:
$\sin x = \pm \frac{1}{\sqrt{2}}$

4. Solve for $x$ in the given interval

The basic angle is $\alpha = \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = 45^\circ$.
Since $\sin x$ can be positive or negative, $x$ can be in any of the four quadrants.
$x = 45^\circ, 180^\circ - 45^\circ, 180^\circ + 45^\circ, 360^\circ - 45^\circ$
$x = 45^\circ, 135^\circ, 225^\circ, 315^\circ$