Trigonometric solutions I am not getting any idea, how to remove the 5x or 2x from this question. Please explain the steps to solve this sum :
What is the number of solutions of the equation 
$\tan(5x)=\cot(2x) ,\;\; 0\leq x\leq \frac{\pi}{2}$
Answer was given as :

4

 A: Hint :
$$\displaystyle \frac{1}{\cot(7x)}=\frac{\tan(5x)+\tan(2x)}{1-\tan(2x)\tan(5x)}$$
Now, your problem is equivalent to find the solutions of the following equation :
$$\cot(7x)=0\,;\:0\leq x\leq\frac{\pi}{2}\:\Longrightarrow\:y=\frac{(2n+1)\pi}{2}\,;\,y\in[0,\frac{7\pi}{2}]$$
A: If all you're required to do is simply to find the number of roots in that range, then just graph it accurately. But here's a slightly unconventional solution to actually solving the equation.
You're given $\tan 5x = \cot 2x \implies \tan 5x \tan 2x = 1$
Now consider $\displaystyle \tan 7x = \frac{\tan 5x + \tan 2x}{1 - \tan 5x \tan 2x}$
From the given conditions, the denominator is always zero.
If $\tan 7x$ were finite, then the only way the limit of the ratio can exist is for $(\tan 5x + \tan 2x)$ to be zero as well. But that would imply $\tan 5x = -\tan 2x$, making the denominator $(1+\tan^2 2x)$, which is strictly greater than zero for all $x \in \mathbb{R}$.
Hence the only remaining option is for $\displaystyle \tan 7x = \pm \infty \implies 7x = \pm \frac{\pi}{2} + n\pi \implies x = \frac{(2n \pm 1)\pi}{14}$.
When you work those out, you'll find the only admissible $x$ values in that range are $\displaystyle x = \{\frac{\pi}{14}, \frac{3\pi}{14}, \frac{5\pi}{14}, \frac{\pi}{2}\}$.
A: $$\tan5x=\cot2x=\tan\left(\frac\pi2-2x\right)$$
$$\implies5x=n\pi+\frac\pi2-2x\iff7x=(2n+1)\frac\pi2$$ where $n$ is any integer
$$\implies x=\frac{(2n+1)\pi}{14}$$
We need $$0\le \frac{(2n+1)\pi}{14}\le\frac\pi2\iff 0\le2n+1\le7\iff0\le n\le3$$
A: First of all, use the identities:-
(1) $\tan (\theta) = \dfrac {\sin (\theta)}{\cos (\theta)}$
(2) $ \cot (\theta) = \dfrac {1}{\tan (\theta)}$
To get $\sin (5x) . \sin (2x) – \cos (5x) . \cos (2x) = 0$
Secondly, apply the compound angle formula:-
$\cos (\theta) . \cos (\phi) –\sin (\theta) . \sin (\phi) = \cos (\theta + \phi)$
The result is then, $\cos (7x) = 0$
The above equation initially yields 7x = π/2 
Since $\cos (\theta) = \cos(2π – \theta), 7x = 3π/2$ is also another solution.
The cosine function is periodic with a period of 2π.
∴ 7x = (π/2) + 2π and also 7π/2 are solutions. The process can go on but the rest will give answers beyond the specified range.
Thus x = π/14 or (...) or [...] or π/2
Hence , there are 4 solutions.
