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 _________.

We have to solve the trigonometric equation

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

for all $$\theta$$ lying in the interval $$[0,\,4\pi]$$, add all those solutions to obtain $$S$$ and finally evaluate $$\dfrac{8S}{\pi}\,. $$ Every algebraic detail is shown below.

First we simplify the left-hand side. We recall the algebraic identity

$$a^{4}+b^{4}=(a^{2}+b^{2})^{2}-2a^{2}b^{2}.$$

Taking $$a=\sin\theta$$ and $$b=\cos\theta$$, and also remembering that $$\sin^{2}\theta+\cos^{2}\theta=1$$, we write

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

Substituting this into the original equation gives

$$1-2\sin^{2}\theta\cos^{2}\theta-\sin\theta\cos\theta=0.$$

For easier manipulation we set

$$x=\sin\theta\cos\theta.$$

Immediately the equation becomes the quadratic

$$1-2x^{2}-x=0 \quad\Longrightarrow\quad -2x^{2}-x+1=0.$$

Multiplying each term by $$-1$$ so that the leading coefficient is positive, we get

$$2x^{2}+x-1=0.$$

Now we solve this quadratic equation by the quadratic formula. The standard formula is

$$x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$$

for an equation $$ax^{2}+bx+cx=0$$. Here $$a=2,\;b=1,\;c=-1$$, so

$$x=\dfrac{-1\pm\sqrt{1^{2}-4(2)(-1)}}{2(2)} =\dfrac{-1\pm\sqrt{1+8}}{4} =\dfrac{-1\pm3}{4}.$$

Hence we obtain two possible numerical values for $$x$$:

$$x_{1}=\dfrac{-1+3}{4}=\dfrac{2}{4}=\dfrac12,\qquad x_{2}=\dfrac{-1-3}{4}=\dfrac{-4}{4}=-1.$$

But $$x=\sin\theta\cos\theta$$, and we also recall the double-angle identity

$$\sin\theta\cos\theta=\dfrac12\sin2\theta.$$

Using this identity, each admissible value of $$x$$ translates into a condition on $$\sin2\theta$$:

1. For $$x=\dfrac12$$ we have $$\dfrac12\sin2\theta=\dfrac12\quad\Longrightarrow\quad\sin2\theta=1.$$

2. For $$x=-1$$ we would have $$\dfrac12\sin2\theta=-1\quad\Longrightarrow\quad\sin2\theta=-2.$$

The second possibility is impossible, because $$\sin$$ of any real angle always lies between $$-1$$ and $$1$$. Hence the only viable condition is

$$\sin2\theta=1.$$

The standard general solution of $$\sin\phi=1$$ is

$$\phi=\dfrac{\pi}{2}+2k\pi,\qquad k\in\mathbb{Z}.$$

Putting $$\phi=2\theta$$ gives

$$2\theta=\dfrac{\pi}{2}+2k\pi\quad\Longrightarrow\quad \theta=\dfrac{\pi}{4}+k\pi,\qquad k\in\mathbb{Z}.$$

Now we list every integer $$k$$ that keeps $$\theta$$ inside the required interval $$[0,\,4\pi]$$.

• For $$k=0$$: $$\theta=\dfrac{\pi}{4}.$$

• For $$k=1$$: $$\theta=\dfrac{\pi}{4}+\pi=\dfrac{5\pi}{4}.$$

• For $$k=2$$: $$\theta=\dfrac{\pi}{4}+2\pi=\dfrac{9\pi}{4}.$$

• For $$k=3$$: $$\theta=\dfrac{\pi}{4}+3\pi=\dfrac{13\pi}{4}.$$

• For $$k=4$$: $$\theta=\dfrac{\pi}{4}+4\pi=\dfrac{17\pi}{4}$$, which exceeds $$4\pi$$ and therefore is not included.

So the complete set of solutions in $$[0,\,4\pi]$$ is

$$\theta\in\left\{\dfrac{\pi}{4},\;\dfrac{5\pi}{4},\;\dfrac{9\pi}{4},\;\dfrac{13\pi}{4}\right\}.$$

Let us now add them to find $$S$$.

$$S=\dfrac{\pi}{4}+\dfrac{5\pi}{4}+\dfrac{9\pi}{4}+\dfrac{13\pi}{4} =\dfrac{(1+5+9+13)\pi}{4} =\dfrac{28\pi}{4}=7\pi.$$

The question finally asks for $$\dfrac{8S}{\pi}$$, so we compute

$$\dfrac{8S}{\pi}=\dfrac{8(7\pi)}{\pi}=56.$$

So, the answer is $$56$$.