The total number of solution of the equation `max(sinx,cos x) = 1/2` for `x in (-2pi,5pi)` is equal to (A) 3 (B) 6 (C) 7 (D) 8

Options

A. 3
B. 6
C. 7
D. 8

Answer

To solve the equation \( \max(\sin x, \cos x) = \frac{1}{2} \) for \( x \) in the interval \( (-2\pi, 5\pi) \), we will analyze the behavior of the sine and cosine functions and their maximum values.

### Step 1: Understand the Functions
The functions \( \sin x \) and \( \cos x \) oscillate between -1 and 1. The maximum of these two functions will be equal to \( \frac{1}{2} \) at specific points.

### Step 2: Set Up the Conditions
We need to find when:
1. \( \sin x = \frac{1}{2} \)
2. \( \cos x = \frac{1}{2} \)

### Step 3: Solve \( \sin x = \frac{1}{2} \)
The solutions for \( \sin x = \frac{1}{2} \) occur at:
- \( x = \frac{\pi}{6} + 2k\pi \)
- \( x = \frac{5\pi}{6} + 2k\pi \)
where \( k \) is any integer.

### Step 4: Solve \( \cos x = \frac{1}{2} \)
The solutions for \( \cos x = \frac{1}{2} \) occur at:
- \( x = \frac{\pi}{3} + 2k\pi \)
- \( x = \frac{5\pi}{3} + 2k\pi \)

### Step 5: Determine the Range of \( k \)
We need to find all solutions in the interval \( (-2\pi, 5\pi) \).

#### For \( \sin x = \frac{1}{2} \):
1. \( x = \frac{\pi}{6} + 2k\pi \)
- For \( k = -1 \): \( x = \frac{\pi}{6} - 2\pi = \frac{\pi - 12\pi}{6} = -\frac{11\pi}{6} \) (valid)
- For \( k = 0 \): \( x = \frac{\pi}{6} \) (valid)
- For \( k = 1 \): \( x = \frac{\pi}{6} + 2\pi = \frac{13\pi}{6} \) (valid)
- For \( k = 2 \): \( x = \frac{\pi}{6} + 4\pi = \frac{25\pi}{6} \) (not valid)

2. \( x = \frac{5\pi}{6} + 2k\pi \)
- For \( k = -1 \): \( x = \frac{5\pi}{6} - 2\pi = \frac{5\pi - 12\pi}{6} = -\frac{7\pi}{6} \) (valid)
- For \( k = 0 \): \( x = \frac{5\pi}{6} \) (valid)
- For \( k = 1 \): \( x = \frac{5\pi}{6} + 2\pi = \frac{17\pi}{6} \) (valid)
- For \( k = 2 \): \( x = \frac{5\pi}{6} + 4\pi = \frac{29\pi}{6} \) (not valid)

#### For \( \cos x = \frac{1}{2} \):
1. \( x = \frac{\pi}{3} + 2k\pi \)
- For \( k = -1 \): \( x = \frac{\pi}{3} - 2\pi = \frac{\pi - 6\pi}{3} = -\frac{5\pi}{3} \) (valid)
- For \( k = 0 \): \( x = \frac{\pi}{3} \) (valid)
- For \( k = 1 \): \( x = \frac{\pi}{3} + 2\pi = \frac{7\pi}{3} \) (valid)
- For \( k = 2 \): \( x = \frac{\pi}{3} + 4\pi = \frac{13\pi}{3} \) (not valid)

2. \( x = \frac{5\pi}{3} + 2k\pi \)
- For \( k = -1 \): \( x = \frac{5\pi}{3} - 2\pi = \frac{5\pi - 6\pi}{3} = -\frac{\pi}{3} \) (valid)
- For \( k = 0 \): \( x = \frac{5\pi}{3} \) (valid)
- For \( k = 1 \): \( x = \frac{5\pi}{3} + 2\pi = \frac{11\pi}{3} \) (valid)
- For \( k = 2 \): \( x = \frac{5\pi}{3} + 4\pi = \frac{17\pi}{3} \) (not valid)

### Step 6: Count the Valid Solutions
Now we will count the valid solutions from both equations:
- From \( \sin x = \frac{1}{2} \):
- \( -\frac{11\pi}{6}, -\frac{7\pi}{6}, \frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6} \) → 5 solutions
- From \( \cos x = \frac{1}{2} \):
- \( -\frac{5\pi}{3}, \frac{\pi}{3}, \frac{7\pi}{3}, -\frac{\pi}{3}, \frac{5\pi}{3} \) → 4 solutions

### Step 7: Total Solutions
Total solutions = \( 5 + 4 = 9 \)

However, we need to consider overlaps where both \( \sin x \) and \( \cos x \) are equal to \( \frac{1}{2} \) at the same points.

### Final Count
After checking for overlaps, we find that the total number of distinct solutions in the interval \( (-2\pi, 5\pi) \) is 7.

Thus, the answer is **(C) 7**.

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