The number of $x\in [0,2\pi]$ for which
$\bigg|\sqrt{2\sin^4 x+18\cos^2 x}-\sqrt{2\cos^4 x+18\sin^2 x}\bigg|=1$ is
What I did try was:
Let $$f(x)=\bigg|\sqrt{2\sin^4 x+18\cos^2 x}-\sqrt{2\cos^4 x+18\sin^2 x}\bigg|$$
then $$f\bigg(\frac{\pi}{2}+x\bigg)=\bigg|\sqrt{2\sin^4 x+18\cos^2 x}-\sqrt{2\cos^4 x+18\sin^2 x}\bigg|$$
So, $\displaystyle \frac{\pi}{2}$ is a time period of that function.
How do I solve this? Pls, I need help.
Best Answer
Let $\sin^2x\cos^2x=t$.
Thus, we have $$2(\sin^4x+\cos^4x)+18(\sin^2x+\cos^2x)-4\sqrt{(\sin^4x+9\cos^2x)(\cos^4x+9\sin^2x)}=1$$ or $$2-4t+18-1=4\sqrt{t^2+9(1-3t)+81t}$$ or $$19-4t=4\sqrt{t^2+54t+9}$$ or $$t=\frac{217}{1016}$$ or $$\sin^22x=\frac{217}{254}$$ or $$1-\cos4x=\frac{217}{127}$$ or $$\cos4x=-\frac{90}{127}.$$ Can you end it now?
I got $8$ roots.
Indeed, $$4x=\pm\left(\pi-\arccos\frac{90}{127}\right)+2\pi k,$$ where $k\in\mathbb Z$ or $$x=\pm\left(\frac{\pi}{4}-\frac{1}{4}\arccos\frac{90}{127}\right)+\frac{\pi k}{2}$$ and since $0\leq x\leq2\pi,$ we obtain: $$0\leq\left(\frac{\pi}{4}-\frac{1}{4}\arccos\frac{90}{127}\right)+\frac{\pi k}{2}\leq2\pi,$$ which gives $$-0.375...-\frac{1}{2}+\frac{1}{2\pi}\arccos\frac{90}{127}\leq k\leq \frac{7}{2}+\frac{1}{2\pi}\arccos\frac{90}{127}=3.624...,$$ for which $k\in\{0,1,2,3\}$ are valid or $$0\leq-\left(\frac{\pi}{4}-\frac{1}{4}\arccos\frac{90}{127}\right)+\frac{\pi k}{2}\leq2\pi,$$ which gives $$0.375...=\frac{1}{2}-\frac{1}{2\pi}\arccos\frac{90}{127}\leq k\leq \frac{9}{2}-\frac{1}{2\pi}\arccos\frac{90}{127}=4.375...,$$ for which $k\in\{1,2,3,4\}$ are valid and we obtain $8$ roots only: $$x_1=\frac{1}{4}\left(\pi+\arccos\frac{90}{127}\right),$$ $$x_2=\frac{1}{4}\left(3\pi+\arccos\frac{90}{127}\right),$$ $$x_3=\frac{1}{4}\left(5\pi+\arccos\frac{90}{127}\right),$$ $$x_4=\frac{1}{4}\left(7\pi+\arccos\frac{90}{127}\right),$$ $$x_5=\frac{1}{4}\left(\pi-\arccos\frac{90}{127}\right),$$ $$x_6=\frac{1}{4}\left(3\pi-\arccos\frac{90}{127}\right),$$ $$x_7=\frac{1}{4}\left(5\pi-\arccos\frac{90}{127}\right)$$ and $$x_8=\frac{1}{4}\left(7\pi-\arccos\frac{90}{127}\right).$$