To solve the inequality $\tan x + 2 \sin x > 3x$, let's analyze both sides and examine where the inequality holds true.

Steps for Solving

1. Domain Consideration:

   • The functions $\tan x$ and $\sin x$ are periodic and have certain restrictions on their domains. Specifically, $\tan x$ is undefined at $x = \frac{\pi}{2} + n\pi$ for integer $n$. So, we’ll need to consider intervals between these points where $\tan x$ is defined.

2. Compare Both Sides:

   • Since this is an inequality involving $\tan x$, $\sin x$, and $x$, it’s challenging to find an exact algebraic solution. However, you can analyze the inequality graphically or numerically by evaluating where $\tan x + 2 \sin x$ is greater than $3x$ within the domain of $x$.

3. Graphical Approach:

   • Graph the functions $f(x) = \tan x + 2 \sin x$ and $g(x) = 3x$.
   • Look for intervals where $f(x) > g(x)$, which would give the solutions to the inequality.

4. Approximate Values:

   • If needed, numerical methods or specific values can be used to approximate where $f(x) > g(x)$.

Would you like a detailed solution using a graphical or numerical approach to find approximate intervals?

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Related Questions:

1. How does the periodicity of trigonometric functions affect inequalities?
2. What methods exist for solving inequalities involving trigonometric and linear terms?
3. Can transformations help simplify inequalities like $\tan x + 2 \sin x > 3x$?
4. How can the properties of the sine and tangent functions be used in inequality proofs?
5. What are typical domains for different trigonometric functions?

Tip: When working with trigonometric inequalities, analyzing domains and using graphing can be especially helpful for visualizing solutions.