Solve the given trigonometric equation using proper identities: $\tan 3x=\tan 5x$

- Hint: When an equation is given in terms of Sine, Cosine, tangent, we must use any of the trigonometric identities to make the equation solvable. There are many inter-relations between Sine, Cosine, tan, secant. These are inter-relations called as identities. Whenever you see conditions such that \[\theta \in R\] , that means inequality is true for all angles. So, directly think of identity which will make your work easy. Use: - $\tan x=\dfrac{\sin x}{\cos x},\sin A-\sin B=2\sin \left( \dfrac{A-B}{2} \right)\cos \left( \dfrac{A+B}{2} \right)$ .


Complete step-by-step solution -

An equality with Sine, Cosine or tangent in them is called trigonometric equality. These are solved by some inter-relations known beforehand.
All the inter-relations which relate Sine, Cosine, tangent, Cotangent, Secant, Cosecant are called trigonometric identities. These trigonometric identities solve the equation and make them simpler to understand for a proof. These are the main and crucial steps to take us nearer to result.
Given equation in the question which we need to solve:
$\tan 3x=\tan 5x$
By basic trigonometry, we can write $\tan x$ in terms of $\sin x,\cos x$ as:
$\tan x=\dfrac{\sin x}{\cos x}$ . By substituting this to our given equation it turns in to:
$\Rightarrow \dfrac{\sin 3x}{\cos 3x}=\dfrac{\sin 5x}{\cos 5x}$
By cross multiplying the terms in the equation, we get it as:
$\sin 3x\cos 5x=\sin 5x\cos 3x$
Subtracting both sides with $\sin 3x\cos 5x$ and dividing it with $\cos 3x\cos 5x$ :
$\Rightarrow \dfrac{-\sin 3x\cos 5x+\sin 5x\cos 3x}{\cos 3x\cos 5x}=0$
We know $\sin A\cos B-\cos A\sin B=\sin \left( A-B \right)$
By using this we can write the above equation as:
$\Rightarrow \dfrac{\sin \left( 5x-3x \right)}{\cos 3x\cos 5x}=0$
By simplifying the above equation, we get the equation as:
$\Rightarrow \dfrac{\sin 2x}{\cos 3x\cos 5x}=0$
So, we get solutions of $\sin 2x=0$ $\cos 3x\cos 5x\ne 0$
Case 1: $\sin 2x=0$
By applying ${{\sin }^{-1}}$ on both sides we get the x values to be
$\Rightarrow 2x=n\pi ;n\in I$
By dividing with 2 on both sides, we get x values to be:
$\Rightarrow x=\dfrac{n\pi }{2};n\in I$
Case 2: $\cos 3x\cos 5x\ne 0$
So, if x is an odd multiple of $\dfrac{\pi }{2}$ $\cos $ will vanish.
So, $x\ne \dfrac{n\pi }{2}$ , n is odd.
So, by taking intersection of both solutions, we get:
$x=\dfrac{n\pi }{2}$ , n is even \[\Rightarrow n=2m\]
By this we get \[x=m\pi ,m\in I\]
Therefore, \[m\pi \] is a solution for a given expression.

Note: The minus sign in \[\sin \left( A-B \right)\] does not matter because the equation is equated to 0. So, we can multiply \[-1\] to get our required form of \[5x-3x\]. Don’t forget to take case 2. Generally, students forget to take case 2 and repeat the answer as \[\dfrac{n\pi }{2}\] but you must take case 2 and the result will be \[m\pi \] . We also know at \[\dfrac{n\pi }{2}\] if n is odd then \[\tan \] is not defined.