The number of solution of `sqrt3Cos^2x=(sqrt3-1)Cosx+1 , x in [0,pi/2]`

Answer

To solve the equation \( \sqrt{3} \cos^2 x = (\sqrt{3} - 1) \cos x + 1 \) for \( x \) in the interval \( [0, \frac{\pi}{2}] \), we will follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the equation to bring all terms to one side:

\[ \sqrt{3} \cos^2 x - (\sqrt{3} - 1) \cos x - 1 = 0 \]

### Step 2: Substituting \( t = \cos x \)
Let \( t = \cos x \). Then, \( \cos^2 x = t^2 \). Substituting this into the equation gives us:

\[ \sqrt{3} t^2 - (\sqrt{3} - 1) t - 1 = 0 \]

### Step 3: Rearranging the Quadratic Equation
Now we can rearrange this equation:

\[ \sqrt{3} t^2 - (\sqrt{3} - 1) t - 1 = 0 \]

### Step 4: Applying the Quadratic Formula
We can solve this quadratic equation using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = \sqrt{3} \), \( b = -(\sqrt{3} - 1) \), and \( c = -1 \).

Calculating the discriminant:

\[ b^2 - 4ac = (-(\sqrt{3} - 1))^2 - 4(\sqrt{3})(-1) \]
\[ = (\sqrt{3} - 1)^2 + 4\sqrt{3} \]
\[ = 3 - 2\sqrt{3} + 1 + 4\sqrt{3} \]
\[ = 4 + 2\sqrt{3} \]

Now substituting back into the quadratic formula:

\[ t = \frac{(\sqrt{3} - 1) \pm \sqrt{4 + 2\sqrt{3}}}{2\sqrt{3}} \]

### Step 5: Finding the Roots
We will calculate the two possible values of \( t \):

1. \( t_1 = \frac{(\sqrt{3} - 1) + \sqrt{4 + 2\sqrt{3}}}{2\sqrt{3}} \)
2. \( t_2 = \frac{(\sqrt{3} - 1) - \sqrt{4 + 2\sqrt{3}}}{2\sqrt{3}} \)

### Step 6: Analyzing the Roots
Next, we need to check if these values of \( t \) (which represent \( \cos x \)) lie within the valid range for \( \cos x \), which is \( [0, 1] \).

1. **For \( t_1 \)**: Check if \( t_1 \) is in \( [0, 1] \).
2. **For \( t_2 \)**: Check if \( t_2 \) is in \( [0, 1] \).

### Step 7: Validating the Solutions
- \( t_1 \) will yield a valid solution if it is in the range \( [0, 1] \).
- \( t_2 \) will yield a valid solution if it is in the range \( [0, 1] \).

### Step 8: Finding \( x \)
If \( t = 1 \), then \( \cos x = 1 \) implies \( x = 0 \).

If \( t = -\frac{1}{\sqrt{3}} \), it is not valid in the interval \( [0, \frac{\pi}{2}] \) since cosine values are non-negative in this range.

### Conclusion
Thus, the only solution we find in the interval \( [0, \frac{\pi}{2}] \) is \( x = 0 \).

### Final Answer
The number of solutions is **1**.

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