Heuristic method for solving $\sin^6 x+\cos^6 x=\dfrac{1}{4}$

Heuristic method for solving $\sin^6 x+\cos^6 x=\dfrac{1}{4}$:

Method III:

Have you ever amazed and marveled by the powerful heuristic skill (the use of substitution, like we used in method I in previous post(Vietnamese Mathematical Olympiad (Trigonometric) Problem of 1962 )) that aimed to help in trick our mind (in a good way) to see the whole problem more clearly and so we could simplify things further to ease our work?

You would be even more staggered by what I am going to reveal to you on this blog post, I will show you how we could develop an already helpful and pretty heuristic thinking!

Borrow from what we have already started in method I, i.e. we let $a=\sin^2 x$ and $b=\cos^2 x$.

We now see that both $a$ and $b$ are non-negative figure with sum of $1$.

We may apply the substitution again, but this one time we let $a=\dfrac{1}{2}-p$ and $b=\dfrac{1}{2}+p$.

The original equation  $\sin^6 x+\cos^6 x=\dfrac{1}{4}$ and all the substitution formulas leads us to

 $a^3+b^3=\dfrac{1}{4}$

$\left(\dfrac{1}{2}-p\right)^3+\left(\dfrac{1}{2}+p\right)^3=\dfrac{1}{4}$

$\dfrac{1}{4}+3p^2=\dfrac{1}{4}$

But

 $\dfrac{1}{4}+3p^2\ge \dfrac{1}{4}$ with equality if and only if $p=0$.

Hence, $a=\dfrac{1}{2}\implies \sin^2 x=\dfrac{1}{2}$.

$\therefore \sin x=\pm \dfrac{1}{\sqrt{2}}$ and hence $x=45^{\circ}+180^{\circ}n,\,135^{\circ}+180^{\circ}n$, where $n\in \mathbb{Z}$.

What we notice here is, we could adopt the heuristic substitution skills TWO TIMES in our plan of attack and we need to recognize when would we achieve the equality of an equation if we have, like in this instance, $x+p=x$, we know $p=0$ is a necessity or we won't achieve equality.

I hope you learn something useful from this blog post but always remember that knowing is not enough, we must apply and everything is POSSIBLE!!!