The number of solutions of $ an^{- 1}4x + an^{- 1}6x = \frac{\pi}{6}$, where $- \frac{1}{2\sqrt{6

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The number of solutions of $	an^{- 1}4x + 	an^{- 1}6x = \frac{\pi}{6}$, where $- \frac{1}{2\sqrt{6}} < x < \frac{1}{2\sqrt{6}}$ is equal to

Accepted Answer

$	an^{- 1}4x + 	an^{- 1}6x = \frac{\pi}{6}$$${\Rightarrow 	an^{- 1}\left( \frac{4x + 6x}{1 + 24x^{2}} ight) = \frac{\pi}{6} }{\Rightarrow \frac{10x}{1 - 24x^{2}} = \frac{1}{\sqrt{3}} }{\Rightarrow 24x^{2} + 10\sqrt{3}x - 1 = 0 }{x = \frac{- 10\sqrt{3} \pm \sqrt{300 + 96}}{48} }{x = \frac{\sqrt{396} - 10\sqrt{3}}{48} }$$Only 1 solution in $\left( - \frac{1}{2\sqrt{6}},\frac{1}{2\sqrt{6}} ight)$

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