Question

The number of solution in [0, π/2] of the equation \[\cos 3x \tan 5x = \sin 7x\] is 

विकल्प

• 5

• 7

• 6

• none of these

Answer 1

6
Given:
\[\cos3x \tan5x = \sin7x\]
\[ \Rightarrow \cos (5x - 2x) \tan5x = \sin (5x + 2x)\]
\[ \Rightarrow \tan5x = \frac{\sin (5x + 2x)}{\cos (5x - 2x)}\]
\[ \Rightarrow \tan5x = \frac{\sin5x \cos2x + \cos5x \sin2x}{\cos5x \cos2x + \sin5x \sin2x}\]
\[ \Rightarrow \frac{\sin5x}{\cos5x} = \frac{\sin5x \cos2x + \cos5x \sin2x}{\cos5x cos2x + \sin5x \sin2x}\]
\[ \Rightarrow \sin5x \cos5x \cos2x + \sin^2 5x \sin2x = \sin5x \cos5x \cos2x + \cos^2 5x \sin2x\]
\[ \Rightarrow \sin^2 5x \sin2x = \cos^2 5x \sin2x\]
\[ \Rightarrow ( \sin^2 5x - \cos^2 5x) \sin2x = 0\]
\[ \Rightarrow (\sin5x - \cos5x) (\sin5x + \cos5x) \sin2x = 0\]
\[\Rightarrow \sin 5 x - \cos 5x = 0 , \sin 5x + \cos 5x = 0\] or \[\sin 2x = 0\] 

\[\Rightarrow \frac{\sin 5x}{\cos 5x} = 1, \frac{\sin 5x}{\cos 5x} = - 1\]

\[\sin 2x = 0\]

Now, 
\[\tan5x = 1 \]
\[ \Rightarrow \tan5x = \tan\frac{\pi}{4}\]
\[ \Rightarrow 5x = n\pi + \frac{\pi}{4}, n \in Z\]
\[ \Rightarrow x = \frac{n\pi}{5} + \frac{\pi}{20}, n \in Z\]

\[\text{ For }n = 0, 1 \text{ and }2,\text{ the values of x are }\frac{\pi}{20}, \frac{\pi}{4}\text{ and }\frac{9\pi}{20}, \text{ respectively} .\]
Or,
\[\tan5x = 1 \]
\[ \Rightarrow \tan5x = \tan \frac{3\pi}{4}\]
\[ \Rightarrow 5x = n\pi + \frac{3\pi}{4}, n \in Z\]
\[ \Rightarrow x = \frac{n\pi}{5} + \frac{3\pi}{20}, n \in Z\]
\[\text{ For }n = 0\text{ and }1,\text{ the values of x are }\frac{3\pi}{20}\text{ and }\frac{7\pi}{20},\text{ respectively .}\]
And,
\[\sin2x = 0 \]
\[ \Rightarrow \sin2x = \sin 0 \]
\[ \Rightarrow 2x = n\pi , n \in Z\]
\[ \Rightarrow x = \frac{n\pi}{2}, n \in Z\]
For n = 0, the value of x is 0 . 
\[\text{ Also, for the odd multiple of }\frac{\pi}{2}, \tan x\text{ is not defined }.\]
Hence, there are six solutions.