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Solve the equation :$\tan 2x=-\cot (x+\large\frac{\pi}{6})$

$\begin{array}{1 1}(A)\;n\pi+ \large\frac{2\pi}{3}&(B)\;n\pi\pm \large\frac{\pi}{6}\\(C)\;n\pi\pm \large\frac{\pi}{8}&(D)\;2n\pi\pm \large\frac{3\pi}{4}\end{array} $

1 Answer

Toolbox:
  • $\tan (\large\frac{\pi}{2}$$+\theta)=-\cot \theta$
  • $\tan \theta=\tan \alpha\Rightarrow \theta=n\pi+\alpha$
Given
$\tan 2x=-\cot (x+\large\frac{\pi}{6})$
$\tan 2x=\tan (\large\frac{\pi}{2}$$+x+\large\frac{\pi}{6})$
$\tan (\large\frac{\pi}{2}$$+\theta)=-\cot \theta$
$\tan 2x=\tan (\large\frac{2\pi}{3}$$+x)$
$2x=n\pi+(\large\frac{2\pi}{3}$$+x) n\in I$
$\tan \theta=\tan \alpha\Rightarrow \theta=n\pi+\alpha$
Hence the required solution is given by
$x=n\pi+ \large\frac{2\pi}{3}$$,n\in I$
Hence (A) is the correct answer.
answered May 22, 2014 by sreemathi.v
 
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