If `n_(1)` denotes the maximum number of roots of `sin theta = k_(1)` in `[0,2pi]` and `n_(2)` denotes the maximum number of roots of `cos theta=k_(2)` in `[0,2pi]`, then

Options

A. `n_(1)+n_(2)=5`
B. `n_(1)+n_(2)=4`
C. `n_(1)+n_(2)=6`
D. `n_(1)+n_(2)=3`

Explanation

To solve the problem, we need to determine the maximum number of roots for the equations \( \sin \theta = k_1 \) and \( \cos \theta = k_2 \) in the interval \([0, 2\pi]\).

### Step 1: Analyze the equation \( \sin \theta = k_1 \)

The sine function oscillates between -1 and 1. Therefore, the maximum number of solutions for \( \sin \theta = k_1 \) depends on the value of \( k_1 \):

- If \( k_1 = 0 \): The equation \( \sin \theta = 0 \) has solutions at \( \theta = 0, \pi, 2\pi \). Thus, there are **3 solutions**.
- If \( k_1 \) is in the range \( (-1, 1) \): The equation will have **2 solutions**.
- If \( k_1 = 1 \) or \( k_1 = -1 \): The equation will have **1 solution** each.

For the maximum number of roots, we consider \( k_1 = 0 \) which gives us \( n_1 = 3 \).

### Step 2: Analyze the equation \( \cos \theta = k_2 \)

Similarly, the cosine function also oscillates between -1 and 1. The maximum number of solutions for \( \cos \theta = k_2 \) also depends on the value of \( k_2 \):

- If \( k_2 = 1 \): The equation \( \cos \theta = 1 \) has the solution \( \theta = 0 \). Thus, there is **1 solution**.
- If \( k_2 = -1 \): The equation \( \cos \theta = -1 \) has the solution \( \theta = \pi \). Thus, there is **1 solution**.
- If \( k_2 \) is in the range \( (-1, 1) \): The equation will have **2 solutions**.

For the maximum number of roots, we consider \( k_2 \) in the range \( (-1, 1) \) which gives us \( n_2 = 2 \).

### Step 3: Calculate \( n_1 + n_2 \)

Now that we have determined \( n_1 \) and \( n_2 \):

- \( n_1 = 3 \)
- \( n_2 = 2 \)

Thus, we can find \( n_1 + n_2 \):

\[ n_1 + n_2 = 3 + 2 = 5 \]

### Final Answer

The value of \( n_1 + n_2 \) is **5**.

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