:Find the parameter m so the equation has real solution

Question

Recently, I have found this problem:

Find the parameter $m \in \Bbb{R}$ so the equation has real solution $$\cos^2(2x)+3m^2=4m(\cos^4(x)-\sin^4(x))$$

I suppose that the answer is $\forall m \in \Bbb{R}$, but how can we show this?

I have tried to factor the right side, obtaining $\cos^2(x)+\sin^2(x)$, but how can we go on?

Answer 1

First we have to factor the right side: $$\cos^2(2x)+3m^2=4m(\cos^2(x)-\sin^2(x))(\cos^2(x)+\sin^2(x))\leftrightarrow \cos^2(2x)+3m^2=4m\cos(2x)$$ Let $t=\cos(2x)$, we have: $$t^2-4mt+3m^2=0$$ Using the quadratic formula, we arrive at: $$t=\frac{4m\pm\sqrt{16m^2-12m^2}}{2}=\frac{4m\pm2m}{2}=3m \; \vee \; m$$ Now, you have to study $\cos(2x)=m$ and $\cos(2x)=3m$ that have real solutions if $m \in [-1,1]$.

Answer 2

One possible solution always is $$x=\frac{1}{2}\cos^{-1}(m)\in \mathbb R$$ for $|m|\leq 1$.

To see that this is a solution, just plug in $x$ into the equation and use that $\cos^4(x)-\sin^4(x)=\cos(2x)$: $$\cos^2(\cos^{-1}(m))+3m^2=4m\cos(\cos^{-1}(m))),$$ that is, $$m^2+3m^2=4m^2$$ which holds for all $m$.