The number of solution in [0, π/2]

Question:

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

(a) 5

(b) 7

(c) 6

(d) none of these

Solution:

(c) 6

Given;

$\cos 3 x \tan 5 x=\sin 7 x$

$\Rightarrow \cos (5 x-2 x) \tan 5 x=\sin (5 x+2 x)$

$\Rightarrow \tan 5 x=\frac{\sin (5 x+2 x)}{\cos (5 x-2 x)}$

$\Rightarrow \tan 5 x=\frac{\sin 5 x \cos 2 x+\cos 5 x \sin 2 x}{\cos 5 x \cos 2 x+\sin 5 x \sin 2 x}$

$\Rightarrow \frac{\sin 5 x}{\cos 5 x}=\frac{\sin 5 x \cos 2 x+\cos 5 x \sin 2 x}{\cos 5 x \cos 2 x+\sin 5 x \sin 2 x}$

$\Rightarrow \sin 5 x \cos 5 x \cos 2 x+\sin ^{2} 5 x \sin 2 x=\sin 5 x \cos 5 x \cos 2 x+\cos ^{2} 5 x \sin 2 x$

$\Rightarrow \sin ^{2} 5 x \sin 2 x=\cos ^{2} 5 x \sin 2 x$

$\Rightarrow\left(\sin ^{2} 5 x-\cos ^{2} 5 x\right) \sin 2 x=0$

$\Rightarrow(\sin 5 x-\cos 5 x)(\sin 5 x+\cos 5 x) \sin 2 x=0$

$\Rightarrow \sin 5 x-\cos 5 x=0, \sin 5 x+\cos 5 x=0$ 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,

$\tan 5 x=1$

$\Rightarrow \tan 5 x=\tan \frac{\pi}{4}$

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

$\Rightarrow x=\frac{\mathrm{n} \pi}{5}+\frac{\pi}{20}, n \in Z$

$F$ or $n=0,1$ and 2 , the value $s$ of $x$ are $\frac{\pi}{20}, \frac{\pi}{4}$ and $\frac{9 \pi}{20}$, respectively.

Or,

$\tan 5 x=1$

$\Rightarrow \tan 5 x=\tan \frac{3 \pi}{4}$

$\Rightarrow 5 x=n \pi+\frac{3 \pi}{4}, n \in Z$

$\Rightarrow x=\frac{\mathrm{n} \pi}{5}+\frac{3 \pi}{20}, n \in Z$

$F$ or $n=0$ and 1, the value $s$ of $x$ are $\frac{3 \pi}{20}$ and $\frac{7 \pi}{20}$, respectively.

And,

$\sin 2 x=0$

$\Rightarrow \sin 2 x=\sin 0$

$\Rightarrow 2 \mathrm{x}=\mathrm{n} \pi, \quad n \in Z$

$\Rightarrow x=\frac{n \pi}{2}, \quad n \in Z$

$F$ or $n=0$, the value of $x$ is $0 .$

Also, $f$ or the odd multiple of $\frac{\pi}{2}, \tan x$ is not defined.

Hence, there are six solutions.

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