Find how many number of roots of the equation `1+ sin ^(3)theta = cos ^(6) theta` in the inteerval `[0,2pi].`

Answer

To solve the equation \(1 + \sin^3 \theta = \cos^6 \theta\) in the interval \([0, 2\pi]\), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ 1 + \sin^3 \theta = \cos^6 \theta \]
We can rearrange it to:
\[ \cos^6 \theta - \sin^3 \theta = 1 \]

### Step 2: Analyze the ranges of the functions
Next, we need to analyze the ranges of \(\cos^6 \theta\) and \(\sin^3 \theta\):
- The range of \(\cos^6 \theta\) is from \(0\) to \(1\) since \(\cos \theta\) varies from \(-1\) to \(1\).
- The range of \(\sin^3 \theta\) is from \(-1\) to \(1\) since \(\sin \theta\) varies from \(-1\) to \(1\).

### Step 3: Determine the maximum and minimum values
From the rearranged equation \(\cos^6 \theta - \sin^3 \theta = 1\), we can analyze the maximum and minimum values:
- The maximum value of \(\cos^6 \theta\) is \(1\) (when \(\theta = 0, \pi, 2\pi\)).
- The minimum value of \(\sin^3 \theta\) is \(-1\) (when \(\theta = \frac{3\pi}{2}\)).

### Step 4: Identify potential roots
We will check specific values of \(\theta\):
- At \(\theta = 0\):
\[ \cos^6(0) - \sin^3(0) = 1 - 0 = 1 \quad \text{(root)} \]
- At \(\theta = \pi\):
\[ \cos^6(\pi) - \sin^3(\pi) = 1 - 0 = 1 \quad \text{(root)} \]
- At \(\theta = 2\pi\):
\[ \cos^6(2\pi) - \sin^3(2\pi) = 1 - 0 = 1 \quad \text{(root)} \]
- At \(\theta = \frac{3\pi}{2}\):
\[ \cos^6\left(\frac{3\pi}{2}\right) - \sin^3\left(\frac{3\pi}{2}\right) = 0 - (-1) = 1 \quad \text{(root)} \]

### Step 5: Check intervals for additional roots
Now we check the intervals:
- Between \(0\) and \(\pi\), and \(\pi\) to \(\frac{3\pi}{2}\), and \(\frac{3\pi}{2}\) to \(2\pi\):
- In these intervals, \(\cos^6 \theta\) is decreasing and \(\sin^3 \theta\) is increasing or vice versa, leading to no additional intersections.

### Conclusion
Thus, the total number of roots in the interval \([0, 2\pi]\) is:
\[ \boxed{4} \]