Question

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

Options

• 5

• 7

• 6

• none of these

Answer 1

6
Given:
$$\cos 3x\tan 5x=\sin 7x$$
$$\Rightarrow \cos (5x-2x)\tan 5x=\sin (5x+2x)$$
$$\Rightarrow \tan 5x=\frac{\sin (5x+2x)}{\cos (5x-2x)}$$
$$\Rightarrow \tan 5x=\frac{\sin 5x\cos 2x+\cos 5x\sin 2x}{\cos 5x\cos 2x+\sin 5x\sin 2x}$$
$$\Rightarrow \frac{\sin 5x}{\cos 5x}=\frac{\sin 5x\cos 2x+\cos 5x\sin 2x}{\cos 5xcos2x+\sin 5x\sin 2x}$$
$$\Rightarrow \sin 5x\cos 5x\cos 2x+{\sin }^{2}5x\sin 2x=\sin 5x\cos 5x\cos 2x+{\cos }^{2}5x\sin 2x$$
$$\Rightarrow {\sin }^{2}5x\sin 2x={\cos }^{2}5x\sin 2x$$
$$\Rightarrow ({\sin }^{2}5x-{\cos }^{2}5x)\sin 2x=0$$
$$\Rightarrow (\sin 5x-\cos 5x)(\sin 5x+\cos 5x)\sin 2x=0$$
$$\Rightarrow \sin 5x-\cos 5x=0,\sin 5x+\cos 5x=0$$ or $$\sin 2x=0$$ 

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

$$\sin 2x=0$$

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

$$\text{ For }n=0,1\text{ and }2,\text{ the values of x are }\frac{\pi }{20},\frac{\pi }{4}\text{ and }\frac{9\pi }{20},\text{ respectively}.$$
Or,
$$\tan 5x=1$$
$$\Rightarrow \tan 5x=\tan \frac{3\pi }{4}$$
$$\Rightarrow 5x=n\pi +\frac{3\pi }{4},n\in Z$$
$$\Rightarrow x=\frac{n\pi }{5}+\frac{3\pi }{20},n\in Z$$
$$\text{ For }n=0\text{ and }1,\text{ the values of x are }\frac{3\pi }{20}\text{ and }\frac{7\pi }{20},\text{ respectively .}$$
And,
$$\sin 2x=0$$
$$\Rightarrow \sin 2x=\sin 0$$
$$\Rightarrow 2x=n\pi ,n\in Z$$
$$\Rightarrow x=\frac{n\pi }{2},n\in Z$$
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.