Question

Range of $\frac{\sin x}{\sin 3x}$

Answer 1

$y=\frac{sinx}{\sin 3x}=\frac{sinx}{3sinx-4{\sin }^{3}x}=\frac{1}{3-4{\sin }^{2}x}$

$\therefore$ $y=\frac{1}{3-4{\sin }^{2}x}$

$\Rightarrow 3y-4y{\sin }^{2}x=1$

$\Rightarrow {\sin }^{2}x=(3y-1)/4y$

$\Rightarrow sinx=\sqrt{\frac{3y-1}{4y}}$

the above equality holds for

$\frac{3y-1}{4y}>0$

$\Rightarrow \frac{(3y-1)\left(4y\right)}{{\left(4y\right)}^2}>0$

$\Rightarrow (3y-1)\left(4y\right)>0$

$\therefore$ $yϵ(-\infty ,0)\cup (1/3,\infty )$

$\therefore$ Range of $\frac{sinx}{\sin 3x}$ is $(-\infty ,0)\cup (\frac{1}{3},\infty )$