Recall that `sinx + cosx =u` (say)
and `sin x cosx =v` (say) are connected by
`(sinx +cosx)^(2) = sin^(2)x + cos^(2)x+2sin cosx`
`rArr u^(2) = 1+2v`
`rArr v=(u^(2)-1)/(2)`
It follows that any rational integral function of `sinx + cosx`, and `sinx cosx` i.e., `R(sinx + cosx, sinx cosx)`, or in our notation R(u,v) can be transformed to `R(u, (u^(2)-1)/2)`. Thus, to solve an equation of the form `R(u,v)=0`, we form a polynomial equation in u and than look for solutions.
The number of solutions of the equation `sin theta + costheta=1 + sintheta costheta` in the interval `[0,4pi]` is

Options

A. Four
B. Six
C. Eight
D. Five

Answer

To solve the equation \( \sin \theta + \cos \theta = 1 + \sin \theta \cos \theta \) in the interval \( [0, 4\pi] \), we can follow these steps:

### Step 1: Define Variables
Let:
\[ u = \sin \theta + \cos \theta \]
\[ v = \sin \theta \cos \theta \]

### Step 2: Use the Identity
From the identity:
\[ u^2 = \sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta \]
we know that:
\[ u^2 = 1 + 2v \]
This implies:
\[ v = \frac{u^2 - 1}{2} \]

### Step 3: Substitute in the Given Equation
The equation can be rewritten using \( u \) and \( v \):
\[ u = 1 + v \]
Substituting \( v \):
\[ u = 1 + \frac{u^2 - 1}{2} \]

### Step 4: Simplify the Equation
Multiply through by 2 to eliminate the fraction:
\[ 2u = 2 + u^2 - 1 \]
Rearranging gives:
\[ u^2 - 2u + 1 = 0 \]

### Step 5: Factor the Quadratic
This can be factored as:
\[ (u - 1)^2 = 0 \]
Thus, we find:
\[ u - 1 = 0 \implies u = 1 \]

### Step 6: Relate Back to Trigonometric Functions
Since \( u = \sin \theta + \cos \theta = 1 \), we can express this as:
\[ \sin \theta + \cos \theta = 1 \]

### Step 7: Solve for \( \theta \)
We can rewrite \( \sin \theta + \cos \theta = 1 \) as:
\[ \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin \theta + \frac{1}{\sqrt{2}} \cos \theta \right) = 1 \]
This simplifies to:
\[ \sin \left( \theta + \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}} \]
The general solutions for \( \sin x = \frac{1}{\sqrt{2}} \) are:
\[ \theta + \frac{\pi}{4} = \frac{\pi}{4} + 2n\pi \quad \text{or} \quad \theta + \frac{\pi}{4} = \frac{3\pi}{4} + 2n\pi \]
Solving these gives:
1. \( \theta = 2n\pi \)
2. \( \theta = 2n\pi + \frac{\pi}{2} \)

### Step 8: Find Solutions in the Interval \( [0, 4\pi] \)
For \( n = 0 \):
- \( \theta = 0 \)
- \( \theta = \frac{\pi}{2} \)

For \( n = 1 \):
- \( \theta = 2\pi \)
- \( \theta = \frac{5\pi}{2} \)

For \( n = 2 \):
- \( \theta = 4\pi \)

### Summary of Solutions
The solutions in the interval \( [0, 4\pi] \) are:
- \( 0 \)
- \( \frac{\pi}{2} \)
- \( 2\pi \)
- \( \frac{5\pi}{2} \)
- \( 4\pi \)

Thus, the total number of solutions is **5**.

### Final Answer
The number of solutions of the equation \( \sin \theta + \cos \theta = 1 + \sin \theta \cos \theta \) in the interval \( [0, 4\pi] \) is **5**.

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