To find the number of solutions of the equation \(3^{2\sec^2 x} + 1 = 10 \cdot 3^{\tan^2 x}\) in the interval \([0, 2\pi]\), we can follow these steps:
### Step 1: Rewrite the equation
We start with the equation:
\[
3^{2\sec^2 x} + 1 = 10 \cdot 3^{\tan^2 x}
\]
Using the identity \(\sec^2 x = 1 + \tan^2 x\), we can rewrite \(3^{2\sec^2 x}\) as:
\[
3^{2(1 + \tan^2 x)} = 3^2 \cdot 3^{2\tan^2 x} = 9 \cdot 3^{2\tan^2 x}
\]
Thus, the equation becomes:
\[
9 \cdot 3^{2\tan^2 x} + 1 = 10 \cdot 3^{\tan^2 x}
\]
### Step 2: Substitute \(t = 3^{\tan^2 x}\)
Let \(t = 3^{\tan^2 x}\). Then, \(3^{2\tan^2 x} = t^2\). Substituting this into the equation gives:
\[
9t^2 + 1 = 10t
\]
### Step 3: Rearrange the equation
Rearranging the equation, we have:
\[
9t^2 - 10t + 1 = 0
\]
### Step 4: Solve the quadratic equation
We can use the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 9\), \(b = -10\), and \(c = 1\):
\[
t = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 9 \cdot 1}}{2 \cdot 9}
\]
Calculating the discriminant:
\[
b^2 - 4ac = 100 - 36 = 64
\]
Now substituting back into the formula:
\[
t = \frac{10 \pm 8}{18}
\]
This gives us two solutions:
\[
t_1 = \frac{18}{18} = 1 \quad \text{and} \quad t_2 = \frac{2}{18} = \frac{1}{9}
\]
### Step 5: Find \(\tan^2 x\) for each \(t\)
1. For \(t_1 = 1\):
\[
3^{\tan^2 x} = 1 \implies \tan^2 x = 0 \implies \tan x = 0
\]
This occurs at:
\[
x = n\pi \quad (n \in \mathbb{Z})
\]
In the interval \([0, 2\pi]\), the solutions are \(x = 0, \pi, 2\pi\).
2. For \(t_2 = \frac{1}{9}\):
\[
3^{\tan^2 x} = \frac{1}{9} \implies \tan^2 x = -2
\]
Since \(\tan^2 x\) cannot be negative, there are no solutions from this case.
### Step 6: Count the number of solutions
From the first case, we found three solutions: \(x = 0\), \(x = \pi\), and \(x = 2\pi\). Therefore, the total number of solutions in the interval \([0, 2\pi]\) is:
\[
\text{Number of solutions} = 3
\]
### Final Answer
The number of solutions of the equation in the interval \([0, 2\pi]\) is \(3\).
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