[inlmath]\text{tg }x\left(1-\text{tg}^2x\right)\left(1-3\text{tg}^2x\right)\left(1+\text{tg }2x\cdot\text{tg }3x\right)>0[/inlmath]
Sad sređujemo deo po deo:
[dispmath]\text{tg }x=\frac{\sin x}{\cos x}\;\Rightarrow x\;\ne\frac{\pi}{2}+k\pi\\
1-\text{tg}^2x=\frac{\cos^2x-\sin^2x}{\cos^2x}\\
1-3\text{tg}^2x=\frac{\cos^2x-3\sin^2x}{\cos^2x}\\
\begin{align}
\phantom{aaaaaaa}1+\text{tg }2x\cdot\text{tg }3x&=1+\frac{2\sin x\cos x}{\cos^2x-\sin^2x}\cdot\frac{\sin\left(2x+x\right)}{\cos\left(2x+x\right)}\\
&=1+\frac{2\sin x\cos x}{\cos^2x-\sin^2x}\cdot\frac{\sin2x\cos x+\cos2x\sin x}{\cos2x\cos x-\sin2x\sin x}\\
&=1+\frac{2\sin x\cos x}{\cos^2x-\sin^2x}\cdot\frac{2\sin x\cos^2x+\cos^2x\sin x-\sin^3x}{\cos^3x-\sin^2x\cos x-2\sin^2x\cos x}\\
&=1+\frac{2\sin x\cancel{\cos x}}{\cos^2x-\sin^2x}\cdot\frac{\sin x\left(3\cos^2x-\sin^2x\right)}{\cancel{\cos x}\left(\cos^2x-3\sin^2x\right)}\\
&=1+\frac{2\sin x\cos x}{\cos^2x-\sin^2x}\cdot\frac{\sin x\left(3\cos^2x-\sin^2x\right)}{\cos^2x-3\sin^2x}\\
&=\frac{\left(\cos^2x-\sin^2x\right)\left(\cos^2x-3\sin^2x\right)+2\sin^2x\left(3\cos^2x-\sin^2x\right)}{\left(\cos^2x-\sin^2x\right)\left(\cos^2x-3\sin^2x\right)}\\
&=\frac{\left(2\cos^2x-1\right)\left(4\cos^2x-3\right)+\left(2-2\cos^2x\right)\left(4\cos^2x-1\right)}{\left(\cos^2x-\sin^2x\right)\left(\cos^2x-3\sin^2x\right)}\\
&=\frac{\cancel{8\cos^4x}-\cancel{6\cos^2x}-\cancel{4\cos^2x}+3-\cancel{8\cos^2x}-2-\cancel{8\cos^4x}+\cancel{2\cos^2x}}{\left(\cos^2x-\sin^2x\right)\left(\cos^2x-3\sin^2x\right)}\\
&=\frac{1}{\left(\cos^2x-\sin^2x\right)\left(\cos^2x-3\sin^2x\right)}
\end{align}[/dispmath]
Ovde isključujemo
[dispmath]x\ne\frac{\pi}{4}+\frac{k\pi}{2}\\
x\ne\frac{\pi}{6}+k\pi\\
x\ne\frac{5\pi}{6}+k\pi[/dispmath]
Sad sve sređene izraze uvrstimo u početni izraz
[dispmath]\text{tg }x\left(1-\text{tg}^2x\right)\left(1-3\text{tg}^2x\right)\left(1+\text{tg }2x\cdot\text{tg }3x\right)>0\iff\\
\iff\text{tg }x\cdot\frac{\cancel{\cos^2x-\sin^2x}}{\cos^2x}\cdot\frac{\cancel{\cos^2x-3\sin^2x}}{\cos^2x}\cdot\frac{1}{\cancel{\left(\cos^2x-\sin^2x\right)}\cancel{\left(\cos^2x-3\sin^2x\right)}}>0\\\
\\\
\Rightarrow\;\frac{\text{tg }x}{\cos^4x}>0\\
\text{tg }x>0[/dispmath]
Napravio sam grafik zadate funkcije u intervalu [inlmath]\left[-\pi,2\pi\right][/inlmath]
- tgx (2).png (23.25 KiB) Pogledano 2057 puta
[dispmath]\text{tg }x>0\;\Rightarrow\;x\in\left(-\pi,-\frac{5\pi}{6}\right)\cup\left(-\frac{5\pi}{6},-\frac{3\pi}{4}\right)\cup\left(-\frac{3\pi}{4},-\frac{\pi}{2}\right)\cup\\
\cup\left(0,\frac{\pi}{6}\right)\cup\left(\frac{\pi}{6},\frac{\pi}{4}\right)\cup\left(\frac{\pi}{4},\frac{\pi}{2}\right)\cup\left(\pi,\frac{7\pi}{6}\right)\cup\left(\frac{7\pi}{6},\frac{5\pi}{4}\right)\cup\left(\frac{5\pi}{4},\frac{3\pi}{2}\right)[/dispmath]
Sva ponuđena rešenja osim [inlmath]\left(B\right)\left(\frac{\pi}{3},\frac{\pi}{2}\right)[/inlmath] koje se nalazi unutar intervala [inlmath]\left(\frac{\pi}{4},\frac{\pi}{2}\right)[/inlmath] otpadaju zbog uslova koje smo dobili za vreme sređivanja izraza.