Trigonometric solutions

Question

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

Answer 1

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 : $$\cot(7x)=0\,;\:0\leq x\leq\frac{\pi}{2}\:\Longrightarrow\:y=\frac{(2n+1)\pi}{2}\,;\,y\in[0,\frac{7\pi}{2}]$$

Answer 2

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 $\displaystyle x = \{\frac{\pi}{14}, \frac{3\pi}{14}, \frac{5\pi}{14}, \frac{\pi}{2}\}$.

Answer 3

$$\tan5x=\cot2x=\tan\left(\frac\pi2-2x\right)$$

$$\implies5x=n\pi+\frac\pi2-2x\iff7x=(2n+1)\frac\pi2$$ where $n$ is any integer

$$\implies x=\frac{(2n+1)\pi}{14}$$

We need $$0\le \frac{(2n+1)\pi}{14}\le\frac\pi2\iff 0\le2n+1\le7\iff0\le n\le3$$

Answer 4

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.