For the equation `cos^(-1)x+cos^(-1)2x+pi=0` , the number of real solution is

Options

A. 1
B. 2
C. 0
D. `oo`

Explanation

To solve the equation \( \cos^{-1} x + \cos^{-1} 2x + \pi = 0 \), we can follow these steps:

### Step 1: Rearrange the equation
We start by isolating the inverse cosine terms:
\[ \cos^{-1} x + \cos^{-1} 2x = -\pi \]

### Step 2: Analyze the range of the inverse cosine functions
The range of \( \cos^{-1} x \) is from \( 0 \) to \( \pi \) for \( x \) in the interval \([-1, 1]\). Similarly, the range of \( \cos^{-1} 2x \) is also from \( 0 \) to \( \pi \) for \( 2x \) in the interval \([-1, 1]\), which means \( x \) must be in the interval \([-0.5, 0.5]\).

### Step 3: Determine the sum of the ranges
Since both \( \cos^{-1} x \) and \( \cos^{-1} 2x \) are non-negative (i.e., they take values from \( 0 \) to \( \pi \)), their sum:
\[ \cos^{-1} x + \cos^{-1} 2x \]
will also be non-negative. Therefore, we can conclude:
\[ \cos^{-1} x + \cos^{-1} 2x \geq 0 \]

### Step 4: Compare with the right side of the equation
The right side of our rearranged equation is \(-\pi\), which is a negative number. Since the left side (the sum of two non-negative numbers) cannot equal a negative number, we conclude that:
\[ \cos^{-1} x + \cos^{-1} 2x \neq -\pi \]

### Step 5: Conclusion on the number of real solutions
Since the left-hand side cannot equal the right-hand side, we conclude that there are no real solutions to the equation:
\[ \text{Number of real solutions} = 0 \]

### Final Answer
The number of real solutions to the equation \( \cos^{-1} x + \cos^{-1} 2x + \pi = 0 \) is **0**.

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