Number of solutions of the equation `(sqrt(3)+1)^(2x)+(sqrt(3)-1)^(2x)=2^(3x)` is________

Explanation

To solve the equation \((\sqrt{3}+1)^{2x} + (\sqrt{3}-1)^{2x} = 2^{3x}\), we will follow these steps:

### Step 1: Rewrite the equation
We start with the original equation:
\[ (\sqrt{3}+1)^{2x} + (\sqrt{3}-1)^{2x} = 2^{3x} \]
We can express \(2^{3x}\) as \(8^x\):
\[ (\sqrt{3}+1)^{2x} + (\sqrt{3}-1)^{2x} = 8^x \]

### Step 2: Divide both sides by \(8^x\)
Next, we divide both sides of the equation by \(8^x\):
\[ \frac{(\sqrt{3}+1)^{2x}}{8^x} + \frac{(\sqrt{3}-1)^{2x}}{8^x} = 1 \]
This simplifies to:
\[ \left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)^{2x} + \left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)^{2x} = 1 \]

### Step 3: Identify trigonometric identities
We know from trigonometric identities that:
\[ \sin 75^\circ = \frac{\sqrt{3}+1}{2\sqrt{2}} \quad \text{and} \quad \cos 75^\circ = \frac{\sqrt{3}-1}{2\sqrt{2}} \]
Thus, we can rewrite our equation as:
\[ (\sin 75^\circ)^{2x} + (\cos 75^\circ)^{2x} = 1 \]

### Step 4: Use the Pythagorean identity
Using the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\), we can see that:
\[ (\sin 75^\circ)^{2x} + (\cos 75^\circ)^{2x} = 1 \]
is satisfied when \(2x = 1\) (since both terms must equal 1 when raised to the power of 1).

### Step 5: Solve for \(x\)
From \(2x = 1\), we find:
\[ x = \frac{1}{2} \]

### Step 6: Determine the number of solutions
Since the equation holds true for \(x = \frac{1}{2}\) and there are no restrictions on \(x\) in the original equation, we conclude that there is exactly **one solution**.

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

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