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Properties Derived from Trigonometric Identities

The number of...

Question

The number of solution in [0, π/2] of the equation $\cos 3 x \tan 5 x = \sin 7 x$ is
(a) 5
(b) 7
(c) 6
(d) none of these

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Solution

(c) 6

Given:

$\cos 3 x \tan 5 x = \sin 7 x \Rightarrow \cos (5 x - 2 x) \tan 5 x = \sin (5 x + 2 x) \Rightarrow \tan 5 x = \frac{\sin (5 x + 2 x)}{\cos (5 x - 2 x)} \Rightarrow \tan 5 x = \frac{\sin 5 x \cos 2 x + \cos 5 x \sin 2 x}{\cos 5 x \cos 2 x + \sin 5 x \sin 2 x} \Rightarrow \frac{\sin 5 x}{\cos 5 x} = \frac{\sin 5 x \cos 2 x + \cos 5 x \sin 2 x}{\cos 5 x c o s 2 x + \sin 5 x \sin 2 x} \Rightarrow \sin 5 x \cos 5 x \cos 2 x + \sin^{2} 5 x \sin 2 x = \sin 5 x \cos 5 x \cos 2 x + \cos^{2} 5 x \sin 2 x \Rightarrow \sin^{2} 5 x \sin 2 x = \cos^{2} 5 x \sin 2 x \Rightarrow (\sin^{2} 5 x - \cos^{2} 5 x) \sin 2 x = 0 \Rightarrow (\sin 5 x - \cos 5 x) (\sin 5 x + \cos 5 x) \sin 2 x = 0$
$\Rightarrow \sin 5 x - \cos 5 x = 0, \sin 5 x + \cos 5 x = 0$ or $\sin 2 x = 0$

$\Rightarrow \frac{\sin 5 x}{\cos 5 x} = 1, \frac{\sin 5 x}{\cos 5 x} = - 1$ or $\sin 2 x = 0$

Now,

$\tan 5 x = 1 \Rightarrow \tan 5 x = \tan \frac{π}{4} \Rightarrow 5 x = n π + \frac{π}{4}, n \in Z \Rightarrow x = \frac{n π}{5} + \frac{π}{20}, n \in Z$

$F or n = 0, 1 and 2, the value s of x are \frac{π}{20}, \frac{π}{4} a n d \frac{9 π}{20}, r e s p e c t i v e l y .$
Or,

$\tan 5 x = 1 \Rightarrow \tan 5 x = \tan \frac{3 π}{4} \Rightarrow 5 x = n π + \frac{3 π}{4}, n \in Z \Rightarrow x = \frac{n π}{5} + \frac{3 π}{20}, n \in Z$

$F or n = 0 and 1, the value s of x are \frac{3 π}{20} a n d \frac{7 π}{20}, r e s p e c t i v e l y .$
And,

$\sin 2 x = 0 \Rightarrow \sin 2 x = \sin 0 \Rightarrow 2 x = n π, n \in Z \Rightarrow x = \frac{n π}{2}, n \in Z$

$F or n = 0, the value of x i s 0 . A l s o, f or the odd multiple of \frac{π}{2}, \tan x is not defined .$

Hence, there are six solutions.

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