The number of solutions of the equations `y=1/3[sin theta+[sin theta +[sin theta]]] and [y+[y]]=2 cos theta ` [ where , [.] denote the greatest integer function ] is/are

Options

A. 0
B. 1
C. 2
D. infinte

Answer

To solve the equations given in the problem, we will analyze each equation step by step.

### Step 1: Analyze the first equation
The first equation is given by:
\[ y = \frac{1}{3} \left[ \sin \theta + [\sin \theta + [\sin \theta]] \right] \]
where \([.]\) denotes the greatest integer function (GIF).

### Step 2: Identify cases for \(\theta\)
We will consider three cases based on the value of \(\theta\):

**Case 1:** \(\theta = \frac{\pi}{2}\)
**Case 2:** \(0 \leq \theta < \frac{\pi}{2}\)
**Case 3:** \(-\frac{\pi}{2} < \theta < 0\)

### Step 3: Calculate \(y\) for each case

**Case 1: \(\theta = \frac{\pi}{2}\)**
\[ \sin \left( \frac{\pi}{2} \right) = 1 \implies y = \frac{1}{3} \left[ 1 + [1 + [1]] \right] = \frac{1}{3} \left[ 1 + [1 + 1] \right] = \frac{1}{3} \left[ 1 + 2 \right] = \frac{1}{3} \times 3 = 1 \]

**Case 2: \(0 \leq \theta < \frac{\pi}{2}\)**
In this range, \(\sin \theta\) varies from 0 to 1. Hence, \([ \sin \theta ] = 0\) (since \(\sin \theta < 1\)).
\[ y = \frac{1}{3} \left[ \sin \theta + [\sin \theta + [\sin \theta]] \right] = \frac{1}{3} \left[ \sin \theta + [\sin \theta + 0] \right] = \frac{1}{3} \left[ \sin \theta + 0 \right] = \frac{1}{3} \sin \theta \]
Since \(\sin \theta\) ranges from 0 to 1, \(y\) will range from 0 to \(\frac{1}{3}\).

**Case 3: \(-\frac{\pi}{2} < \theta < 0\)**
In this range, \(\sin \theta\) varies from -1 to 0. Hence, \([ \sin \theta ] = -1\) (since \(\sin \theta < 0\)).
\[ y = \frac{1}{3} \left[ \sin \theta + [-1 + [-1]] \right] = \frac{1}{3} \left[ \sin \theta - 1 \right] \]
As \(\sin \theta\) varies from -1 to 0, \(y\) will range from \(-\frac{2}{3}\) (when \(\sin \theta = -1\)) to \(-\frac{1}{3}\) (when \(\sin \theta = 0\)).

### Step 4: Analyze the second equation
The second equation is:
\[ [y + [y]] = 2 \cos \theta \]

### Step 5: Substitute \(y\) values into the second equation
We will analyze the values of \(y\) obtained from the first equation in the second equation.

1. **For \(y = 1\)**:
\[ [1 + [1]] = [1 + 1] = [2] = 2 \implies 2 \cos \theta = 2 \implies \cos \theta = 1 \implies \theta = 0 \]
This case is valid.

2. **For \(y\) in the range \([0, \frac{1}{3}]\)**:
\[ [y + [y]] = [y + 0] = [y] \]
Since \(y\) can take values from 0 to \(\frac{1}{3}\), \([y] = 0\) for all values in this range. Thus:
\[ 0 = 2 \cos \theta \implies \cos \theta = 0 \implies \theta = \frac{\pi}{2} \text{ or } \theta = \frac{3\pi}{2} \]
However, \(\theta = \frac{\pi}{2}\) is not included in this case.

3. **For \(y\) in the range \([-1, -\frac{1}{3}]\)**:
\[ [y + [y]] = [y - 1] = [-1] = -1 \]
This implies:
\[ -1 = 2 \cos \theta \implies \cos \theta = -\frac{1}{2} \implies \theta = \frac{2\pi}{3} \text{ or } \theta = \frac{4\pi}{3} \]
Both values are valid.

### Step 6: Conclusion
From the analysis, we have found:
- One solution from \(y = 1\) (where \(\theta = 0\)).
- No valid solutions from the second case.
- Two solutions from the third case (\(\theta = \frac{2\pi}{3}\) and \(\theta = \frac{4\pi}{3}\)).

Thus, the total number of solutions is **three**.