Let $$S$$ be the sum of all solutions (in radians) of the equation $$\sin^4\theta + \cos^4\theta - \sin\theta\cos\theta = 0$$ in $$[0, 4\pi]$$ then $$\frac{8S}{\pi}$$ is equal to _________.

Step 1: Simplify the given equation

Start with the given trigonometric equation:

$$\sin^4 \theta + \cos^4 \theta - \sin \theta \cos \theta = 0$$

Use the standard algebraic identity $$a^2 + b^2 = (a+b)^2 - 2ab$$ to rewrite the first two terms, where $$a = \sin^2 \theta$$ and $$b = \cos^2 \theta$$:

$$\sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta$$

Apply the fundamental trigonometric identity $$\sin^2 \theta + \cos^2 \theta = 1$$:

$$\sin^4 \theta + \cos^4 \theta = (1)^2 - 2\sin^2 \theta \cos^2 \theta$$
$$\sin^4 \theta + \cos^4 \theta = 1 - 2\sin^2 \theta \cos^2 \theta$$

Substitute this simplified expression back into the original equation:

$$1 - 2\sin^2 \theta \cos^2 \theta - \sin \theta \cos \theta = 0$$

Step 2: Solve the quadratic equation

To make the equation easier to solve, let us introduce a temporary variable $$t = \sin \theta \cos \theta$$.

$$1 - 2t^2 - t = 0$$

Rearrange the terms into a standard quadratic format:

$$2t^2 + t - 1 = 0$$

Factorize the quadratic equation:

$$2t^2 + 2t - t - 1 = 0$$
$$2t(t + 1) - 1(t + 1) = 0$$
$$(2t - 1)(t + 1) = 0$$

This gives two possible cases for $$t$$:

• Case 1: $$t = \frac{1}{2} \implies \sin \theta \cos \theta = \frac{1}{2}$$
• Case 2: $$t = -1 \implies \sin \theta \cos \theta = -1$$

Multiply both sides of each case by 2 to utilize the double angle identity $$\sin 2\theta = 2\sin\theta\cos\theta$$:

• From Case 1: $$2\sin \theta \cos \theta = 1 \implies \sin 2\theta = 1$$
• From Case 2: $$2\sin \theta \cos \theta = -2 \implies \sin 2\theta = -2$$

Since the value of any sine function must lie perfectly between $$-1$$ and $$1$$, Case 2 ($$\sin 2\theta = -2$$) is mathematically impossible and is rejected.

Step 3: Find the solutions for the angle

We are left with the valid trigonometric equation:

$$\sin 2\theta = 1$$

The problem specifies the domain for $$\theta$$ is $$[0, 4\pi]$$.

Therefore, the domain for the double angle $$2\theta$$ will be exactly twice that interval, which is $$[0, 8\pi]$$.

We need to find all angles within $$[0, 8\pi]$$ where the sine value is exactly $$1$$. These occur at $$\frac{\pi}{2}$$ and repeat every full cycle of $$2\pi$$:

• $$2\theta = \frac{\pi}{2}$$
• $$2\theta = 2\pi + \frac{\pi}{2} = \frac{5\pi}{2}$$
• $$2\theta = 4\pi + \frac{\pi}{2} = \frac{9\pi}{2}$$
• $$2\theta = 6\pi + \frac{\pi}{2} = \frac{13\pi}{2}$$

Divide each value by 2 to find the individual roots for $$\theta$$:

• $$\theta = \frac{\pi}{4}$$
• $$\theta = \pi + \frac{\pi}{4} = \frac{5\pi}{4}$$
• $$\theta = 2\pi + \frac{\pi}{4} = \frac{9\pi}{4}$$
• $$\theta = 3\pi + \frac{\pi}{4} = \frac{13\pi}{4}$$

Step 4: Calculate the sum and final expression

Let $$S$$ be the sum of all these valid roots:

$$S = \frac{\pi}{4} + \frac{5\pi}{4} + \frac{9\pi}{4} + \frac{13\pi}{4}$$
$$S = \frac{28\pi}{4}$$
$$S = 7\pi$$

The problem asks for the value of the expression $$\frac{8S}{\pi}$$. Substitute our calculated sum $$S$$ into this expression:

$$\text{Value} = \frac{8(7\pi)}{\pi}$$
$$\text{Value} = 8 \times 7$$
$$\text{Value} = 56$$

The final answer is 56.