The number of real solutions `(x, y)`, where `|y| = sin x, y = cos^-1 (cosx),-2pi leq x leq 2pi` is

Options

A. 2
B. 1
C. 3
D. 4

Answer

To find the number of real solutions \((x, y)\) where \(|y| = \sin x\) and \(y = \cos^{-1}(\cos x)\) for \(-2\pi \leq x \leq 2\pi\), we will analyze the equations step by step.

### Step 1: Analyze \(|y| = \sin x\)

The equation \(|y| = \sin x\) implies that \(y\) can take two values:
1. \(y = \sin x\)
2. \(y = -\sin x\)

The sine function oscillates between -1 and 1. Therefore, the possible values for \(y\) are constrained to the interval \([-1, 1]\).

### Step 2: Analyze \(y = \cos^{-1}(\cos x)\)

The function \(y = \cos^{-1}(\cos x)\) gives the principal value of the angle whose cosine is \(\cos x\). The range of \(\cos^{-1}\) is \([0, \pi]\). However, since \(x\) can take values from \(-2\pi\) to \(2\pi\), we need to consider how the function behaves over this interval.

- For \(x\) in \([-2\pi, -\pi]\), \(y = \cos^{-1}(\cos x) = -x\) (which gives values from \(\pi\) to \(2\pi\)).
- For \(x\) in \([- \pi, 0]\), \(y = \cos^{-1}(\cos x) = -x\) (which gives values from \(\pi\) to \(0\)).
- For \(x\) in \([0, \pi]\), \(y = \cos^{-1}(\cos x) = x\) (which gives values from \(0\) to \(\pi\)).
- For \(x\) in \([\pi, 2\pi]\), \(y = \cos^{-1}(\cos x) = 2\pi - x\) (which gives values from \(\pi\) to \(0\)).

### Step 3: Find intersections of the graphs

Now we need to find the intersections of the two graphs:
1. \(y = \sin x\)
2. \(y = \cos^{-1}(\cos x)\)

#### For \(y = \sin x\):
- The graph oscillates between -1 and 1.

#### For \(y = \cos^{-1}(\cos x)\):
- The graph oscillates between 0 and \(\pi\).

### Step 4: Identify the intervals

We will check the intervals:
1. From \(-2\pi\) to \(-\pi\): Here, \(\sin x\) is negative, and \(\cos^{-1}(\cos x)\) is positive. No intersection.
2. From \(-\pi\) to \(0\): \(\sin x\) goes from 0 to -1, and \(\cos^{-1}(\cos x)\) goes from \(\pi\) to 0. No intersection.
3. From \(0\) to \(\pi\): \(\sin x\) goes from 0 to 1, and \(\cos^{-1}(\cos x)\) goes from 0 to \(\pi\). They intersect at \(x = 0\) and \(x = \frac{\pi}{2}\).
4. From \(\pi\) to \(2\pi\): \(\sin x\) goes from 0 to -1, and \(\cos^{-1}(\cos x)\) goes from 0 to \(\pi\). No intersection.

### Step 5: Count the solutions

From our analysis, we found:
- One solution at \(x = 0\)
- One solution at \(x = \frac{\pi}{2}\)

Thus, there are a total of **2 real solutions** \((x, y)\) in the given interval.

### Final Answer
The number of real solutions \((x, y)\) is **2**.

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