Find the number of solution of the equation `sinx=x^(2)+x+1`.

Explanation

To find the number of solutions for the equation \( \sin x = x^2 + x + 1 \), we can follow these steps:

### Step 1: Understand the Functions
We have two functions to consider:
- \( y_1 = \sin x \)
- \( y_2 = x^2 + x + 1 \)

### Step 2: Analyze the Range of \( \sin x \)
The sine function oscillates between -1 and 1 for all real values of \( x \):
\[ -1 \leq \sin x \leq 1 \]

### Step 3: Analyze the Quadratic Function
The quadratic function \( y_2 = x^2 + x + 1 \) is a parabola that opens upwards. To find its minimum value, we can complete the square or use the vertex formula. The vertex \( x \) coordinate is given by:
\[ x = -\frac{b}{2a} = -\frac{1}{2 \cdot 1} = -\frac{1}{2} \]
Substituting \( x = -\frac{1}{2} \) into \( y_2 \):
\[ y_2 = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) + 1 = \frac{1}{4} - \frac{1}{2} + 1 = \frac{3}{4} \]
Thus, the minimum value of \( y_2 \) is \( \frac{3}{4} \).

### Step 4: Compare the Ranges
Since \( y_1 \) (which is \( \sin x \)) can take values from -1 to 1, and \( y_2 \) (which is \( x^2 + x + 1 \)) has a minimum value of \( \frac{3}{4} \), we can conclude:
\[ \sin x \text{ can equal } x^2 + x + 1 \text{ only when } \sin x \geq \frac{3}{4} \]

### Step 5: Find the Values of \( x \)
Now we need to find the values of \( x \) where \( \sin x \geq \frac{3}{4} \). The sine function reaches \( \frac{3}{4} \) at specific angles:
- In the first quadrant, \( x = \arcsin\left(\frac{3}{4}\right) \)
- In the second quadrant, \( x = \pi - \arcsin\left(\frac{3}{4}\right) \)

### Step 6: Determine the Periodicity
Since \( \sin x \) is periodic with a period of \( 2\pi \), we can find additional solutions:
- The solutions in the first and second quadrants will repeat every \( 2\pi \).

### Step 7: Count the Solutions
For each period of \( 2\pi \), there are two solutions (one in the first quadrant and one in the second quadrant). As \( x^2 + x + 1 \) is always greater than or equal to \( \frac{3}{4} \), we can find that there are infinitely many solutions, but we can count the number of intersections in a limited range.

### Conclusion
In the interval \( [0, 2\pi] \), we find that there are exactly 2 solutions. However, since the sine function oscillates indefinitely, the total number of solutions is infinite.

### Final Answer
The number of solutions of the equation \( \sin x = x^2 + x + 1 \) is infinite.