Question

Find which of the number of the form $(n\pi -ta{n}^{-1}3)$, where $nϵl$, are solution for $12tan2x+\frac{\sqrt{10}}{cosx}+1=0$

• A. $n=k,kϵz$
• B. $n=(3k+1),kϵz$
• C. $n=(2k+1),kϵz$
• D. $n=(4k+1),kϵz$

Answer 1

The correct option is C $n=(2k+1),kϵz$

$x=n\pi -{\tan }^{-1}3\Rightarrow \tan x=-3$
Now, $\tan 2x=\frac{2\tan x}{1-{\tan }^{2}x}=\frac{3}{4}$ and $\cos x=\pm \frac{1}{\sqrt{1+\tan x}}=\pm \frac{1}{\sqrt{10}}$
On substituting these values in the given equation,we find only $=\frac{-1}{\sqrt{10}}$ satisfies the equation, so equation true for $=-3$ and $\cos x=-\frac{1}{\sqrt{10}}.$
Which is possible if $x$ lies in $2^{nd}$ quadrant.
So,n must be odd integer