Points of Non-Differentiability of \( f(x) = |\cos x| + 3 \)

Step 1: \( \cos x \) is differentiable everywhere, but \( |\cos x| \) is not differentiable where \( \cos x = 0 \).

Step 2: In the interval \( [-\pi, \pi] \), we have:

\[ \cos x = 0 \Rightarrow x = -\frac{\pi}{2},\ \frac{\pi}{2} \]

So \( f(x) = |\cos x| + 3 \) is not differentiable at these two points due to sharp turns.

✅ Final Answer:   \( \boxed{2 \text{ points}} \)

We are given:

\[ \text{Evaluate } \tan\left(\frac{\pi}{4} + \theta\right) \cdot \tan\left(\frac{3\pi}{4} + \theta\right) \]

✳ Step 1: Use identity

\[ \tan\left(A + B\right) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \] But we don’t need expansion — use known angle values:

\[ \tan\left(\frac{\pi}{4} + \theta\right) = \frac{1 + \tan\theta}{1 - \tan\theta} \]

\[ \tan\left(\frac{3\pi}{4} + \theta\right) = \frac{-1 + \tan\theta}{1 + \tan\theta} \]

✳ Step 2: Multiply

\[ \left(\frac{1 + \tan\theta}{1 - \tan\theta}\right) \cdot \left(\frac{-1 + \tan\theta}{1 + \tan\theta}\right) \]

Simplify:

\[ = \frac{(1 + \tan\theta)(-1 + \tan\theta)}{(1 - \tan\theta)(1 + \tan\theta)} = \frac{(\tan^2\theta - 1)}{1 - \tan^2\theta} = \boxed{-1} \]

✅ Final Answer:

\[ \boxed{-1} \]

Given:

\[ \sin x = \sin y \quad \text{and} \quad \cos x = \cos y \]

✳ Step 1: Use the identity for sine

\[ \sin x = \sin y \Rightarrow x = y + 2n\pi \quad \text{or} \quad x = \pi - y + 2n\pi \]

✳ Step 2: Use the identity for cosine

\[ \cos x = \cos y \Rightarrow x = y + 2m\pi \quad \text{or} \quad x = -y + 2m\pi \]

? Combine both conditions

For both \( \sin x = \sin y \) and \( \cos x = \cos y \) to be true, the only consistent solution is:

\[ x = y + 2n\pi \Rightarrow x - y = 2n\pi \]

✅ Final Answer:

\[ \boxed{x - y = 2n\pi \quad \text{for } n \in \mathbb{Z}} \]