Question

Solve the following equation
$ta{n}^{-1}2x+ta{n}^{-1}3x=\frac{\pi }{4}$

• A. $1$
• B. $\frac{1}{6}$
• C. $\frac{1}{3}$
• D. $-\frac{1}{6}$

Answer 1

Answer: (B)

We have, $ta{n}^{-1}2x+ta{n}^{-1}3x=\frac{\pi }{4}$
$\Rightarrow ta{n}^{-1}\left\{\frac{2x+3x}{1-2x\times 3x}\right\}=ta{n}^{-1}1,$
if $6x^2<1$
$\Rightarrow \frac{5x}{1-6x^2}=1$,
if $6x^2<1$
$\Rightarrow 6x+5x-1=0$ and $x^2<\frac{1}{6}$
$\Rightarrow \left(6x-1\right)\left(x+1\right)=0$ and $-\frac{1}{\sqrt{6}}<x<\frac{1}{\sqrt{6}}$
$\Rightarrow x=-1,\frac{1}{6}$ and $-\frac{1}{\sqrt{6}}<x<\frac{1}{\sqrt{6}}$
$\Rightarrow x=\frac{1}{6}$