The number of the real roots of the equation $2\cos (x(x + 1)) = 2^x + 2^{-x}$ is

We need to find all real values of $x$ where the left side equals the right side of our equation. Let's analyze each side separately to understand their behavior.

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The cosine function always produces values between -1 and 1.

Since $\cos(\text{anything})$ ranges from -1 to 1, when we multiply by 2:

$2\cos(x(x+1))$ ranges from $2(-1) = -2$ to $2(1) = 2$

No matter what value $x(x+1)$ takes, the left side can never exceed 2 or go below -2.

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We'll use the AM-GM inequality here. The AM-GM inequality states that for any two positive numbers $a$ and $b$:

$\tfrac{a + b}{2} \geq \sqrt{ab}$

Both $2^x$ and $2^{-x}$ are always positive (exponential functions with positive bases are always positive).

Let's set $a = 2^x$ and $b = 2^{-x}$:

$\tfrac{2^x + 2^{-x}}{2} \geq \sqrt{2^x \cdot 2^{-x}}$

$\sqrt{2^x \cdot 2^{-x}} = \sqrt{2^{x + (-x)}} = \sqrt{2^0} = \sqrt{1} = 1$

Therefore:

$\tfrac{2^x + 2^{-x}}{2} \geq 1$

$2^x + 2^{-x} \geq 2$

The AM-GM inequality tells us that equality occurs when $a = b$, meaning $2^x = 2^{-x}$.

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For $2^x = 2^{-x}$ to be true:

Taking log of both sides: $x \log(2) = -x \log(2)$

This gives us: $x = -x$

Therefore: $2x = 0$, so $x = 0$

At $x = 0$: $2^0 + 2^{-0} = 1 + 1 = 2$

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Now we know:

Left side: Can range from -2 to 2

Right side: Always $\geq 2$, and equals 2 only when $x = 0$

Since the right side is always at least 2, and the left side can't exceed 2, they can only meet when both sides equal exactly 2.

At $x = 0$:

Left side: $2\cos(0 \cdot 1) = 2\cos(0) = 2(1) = 2$

Right side: $2^0 + 2^{-0} = 1 + 1 = 2$

Both sides equal 2, so $x = 0$ is our solution!

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For any other solution to exist, we would need both sides to equal the same value. But:

When $x \geq 0$, the right side is strictly greater than 2

The left side cannot exceed 2

Therefore, no other intersections are possible.

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The equation $2\cos(x(x+1)) = 2^x + 2^{-x}$ has exactly 1 real root: $x = 0$.

This problem beautifully demonstrates how understanding the ranges of different functions can help us solve complex equations without heavy computation!