What is the general solution of the equation-

(tanx/tan2x)+(tan2x/tanx)+2=0

To solve the equation \(\frac{\tan x}{\tan 2x} + \frac{\tan 2x}{\tan x} + 2 = 0\), we first need to simplify and manipulate the expression for easier analysis. Let's break it down step by step.

Understanding the Equation

The equation consists of terms involving the tangent function and its double angle identity. Recall that the double angle identity for tangent is given by:

• \(\tan 2x = \frac{2 \tan x}{1 - \tan^2 x}\)

This identity will be key in our manipulation of the equation.

Simplifying the Components

Let’s substitute \(\tan 2x\) in the equation:

We can rewrite the original equation in terms of \(\tan x\):

• Let \(y = \tan x\).
• Thus, \(\tan 2x = \frac{2y}{1 - y^2}\).

Now substituting these into our equation, we get:

\(\frac{y}{\frac{2y}{1 - y^2}} + \frac{\frac{2y}{1 - y^2}}{y} + 2 = 0\)

Rearranging the Expression

This simplifies to:

• \(\frac{y(1 - y^2)}{2y} + \frac{2y}{y(1 - y^2)} + 2 = 0\)

Reducing further gives:

• \(\frac{1 - y^2}{2} + \frac{2}{1 - y^2} + 2 = 0\)

Finding a Common Denominator

The next step is to find a common denominator to combine these fractions. The common denominator is \(2(1 - y^2)\):

• Converting each term:
• \(\frac{(1 - y^2)^2 + 4}{2(1 - y^2)} = 0\)

Now, we can set the numerator equal to zero:

\((1 - y^2)^2 + 4 = 0\)

Solving the Quadratic Equation

This leads us to solve:

• \((1 - y^2)^2 = -4\)

Since the square of a real number cannot be negative, we find that there are no real solutions for \(y = \tan x\). Thus, the equation has no solutions among real numbers.

Considering Complex Solutions

If we expand our consideration to complex solutions, we can set:

• Let \(1 - y^2 = 2i\) or \(1 - y^2 = -2i\), leading to complex values for \(y\).

However, for practical purposes and most applications, we typically focus on real solutions in trigonometric equations unless specified otherwise.

Final Thoughts

In summary, the equation \(\frac{\tan x}{\tan 2x} + \frac{\tan 2x}{\tan x} + 2 = 0\) does not yield any real solutions. Understanding such equations often requires a solid grasp of trigonometric identities and complex numbers. If you have further questions or need clarification on any part, feel free to ask!