Question

The number of solutions of the equation \(2\theta - \cos^2(\theta) + \sqrt{2} = 0\) in R is equal to ___________.

Answer 1

Correct Answer: 1

To determine the number of solutions for the equation \(2\theta - \cos^2(\theta) + \sqrt{2} = 0\) in \( \mathbb{R} \), we proceed as follows:

First, we express the equation in terms of \(\cos^2(\theta)\):

\(2\theta + \sqrt{2} = \cos^2(\theta)\)

We know that \(\cos^2(\theta) \in [0, 1]\). Therefore, the expression on the left, \(2\theta + \sqrt{2}\), must also lie within this interval.

Set the inequality:

\(0 \leq 2\theta + \sqrt{2} \leq 1\)

From \(2\theta + \sqrt{2} \geq 0\), we find:

\(\theta \geq -\frac{\sqrt{2}}{2}\)

From \(2\theta + \sqrt{2} \leq 1\), we find:

\(\theta \leq \frac{1-\sqrt{2}}{2}\)

Hence, \(\theta\) must satisfy:

\(-\frac{\sqrt{2}}{2} \leq \theta \leq \frac{1-\sqrt{2}}{2}\)

Further analyze the function \(f(\theta) = 2\theta + \sqrt{2} - \cos^2(\theta)\). Investigating the derivative:

\(f'(\theta) = 2 + \sin(2\theta)\)

Since \(\sin(2\theta)\) oscillates between \(-1\) and \(1\), \(f'(\theta)\) varies in \([1, 3]\), ensuring \(f\) is strictly increasing within any interval given by \(\cos^2(\theta)\)'s validity.

Thus, \(f(\theta)\) has at most one solution in any such interval as \(f(\theta)\) is monotonic. Verify endpoints:

At \(\theta = -\frac{\sqrt{2}}{2}\):

\(f\left(-\frac{\sqrt{2}}{2}\right) = 2(-\frac{\sqrt{2}}{2}) + \sqrt{2} - \cos^2(-\frac{\sqrt{2}}{2}) = 0\)

At \(\theta = \frac{1-\sqrt{2}}{2}\):

\(f\left(\frac{1-\sqrt{2}}{2}\right) = 2(\frac{1-\sqrt{2}}{2}) + \sqrt{2} - \cos^2(\frac{1-\sqrt{2}}{2}) = 1 - \cos^2(\frac{1-\sqrt{2}}{2}) > 0\)

Thus, the only viable solution lies at the endpoint where \(\theta = -\frac{\sqrt{2}}{2}\).

In conclusion, there is exactly one solution within the specified range, consistent with \(1,1\).