MHB Solve the equation cos^8θ+sin^8θ−2(1−cos^2θsin^2θ)^2+1=0

[lfdahl]

Find all possible $\theta$, that satisfy the equation:

$$\cos^8\theta+\sin^8\theta-2(1-\cos^2\theta\sin^2\theta)^2+1 = 0$$

 

[castor28]

Find all possible $\theta$, that satisfy the equation:

$$\cos^8\theta+\sin^8\theta-2(1-\cos^2\theta\sin^2\theta)^2+1 = 0$$

[sp]
Let us write $x=\cos^2\theta$; this implies $\sin^2\theta=1-x$. The equation becomes
$$\begin{align*}
x^4 + (1-x)^4 -2(1 - x(1-x))+1&=0\\
2x^4-4x^3+4x^2-2x&=0\\
2x(x-1)(x^2-x+1)&=0
\end{align*}$$
As $x^2-x+1$ has no real root, the only solutions are $x=0$ and $x=1$ (since $x\ge0$). These correspond to $\cos\theta=0,\,\pm1$ and $\theta=\dfrac{n\pi}{2}$, $n\in\mathbb{Z}$.
[/sp]

 

[Opalg]

[sp]
$$\cos^8\theta+\sin^8\theta-2(1-\cos^2\theta\sin^2\theta)^2+1 = 0$$
$$\cos^8\theta+\sin^8\theta-2\cos^4\theta\sin^4\theta +4\cos^2\theta\sin^2\theta - 1 = 0$$
$$(\cos^4\theta - \sin^4\theta)^2 +4\cos^2\theta\sin^2\theta - 1 = 0$$
$$\bigl((\cos^2\theta - \sin^2\theta)(\cos^2\theta + \sin^2\theta)\bigr)^2 +4\cos^2\theta\sin^2\theta - 1 = 0$$
$$(\cos^2\theta - \sin^2\theta)^2 +4\cos^2\theta\sin^2\theta - 1 = 0$$
$$(\cos^2\theta + \sin^2\theta)^2 - 1 = 0$$
$$1 - 1 = 0$$ That is a tautology, so the equation is true for all $\theta$.

[It looks as though castor28 omitted to square $(1 - x(1-x))$.]
[/sp]

 

[lfdahl]

[sp]
Let us write $x=\cos^2\theta$; this implies $\sin^2\theta=1-x$. The equation becomes
$$\begin{align*}
x^4 + (1-x)^4 -2(1 - x(1-x))+1&=0\\
2x^4-4x^3+4x^2-2x&=0\\
2x(x-1)(x^2-x+1)&=0
\end{align*}$$
As $x^2-x+1$ has no real root, the only solutions are $x=0$ and $x=1$ (since $x\ge0$). These correspond to $\cos\theta=0,\,\pm1$ and $\theta=\dfrac{n\pi}{2}$, $n\in\mathbb{Z}$.
[/sp]

Thankyou, castor28, for your participation!

Spoiler

It seems, Opalg is right: You forgot to square the parenthesis. The correct answer is: $\theta \in \Bbb{R}$.

 

[lfdahl]

[sp]
$$\cos^8\theta+\sin^8\theta-2(1-\cos^2\theta\sin^2\theta)^2+1 = 0$$
$$\cos^8\theta+\sin^8\theta-2\cos^4\theta\sin^4\theta +4\cos^2\theta\sin^2\theta - 1 = 0$$
$$(\cos^4\theta - \sin^4\theta)^2 +4\cos^2\theta\sin^2\theta - 1 = 0$$
$$\bigl((\cos^2\theta - \sin^2\theta)(\cos^2\theta + \sin^2\theta)\bigr)^2 +4\cos^2\theta\sin^2\theta - 1 = 0$$
$$(\cos^2\theta - \sin^2\theta)^2 +4\cos^2\theta\sin^2\theta - 1 = 0$$
$$(\cos^2\theta + \sin^2\theta)^2 - 1 = 0$$
$$1 - 1 = 0$$ That is a tautology, so the equation is true for all $\theta$.

[It looks as though castor28 omitted to square $(1 - x(1-x))$.]
[/sp]

Thankyou, Opalg, for a short and elegant solution!

Spoiler

Yes, the equation is indeed a tautology!