The number of solutions of $$\cos x = |\sin x|$$, such that $$-4\pi \le x \le 4\pi$$ is

We need to find the number of solutions of $$ \cos x = |\sin x| $$ in the interval $$ -4\pi \le x \le 4\pi $$.

Since the RHS is $$ |\sin x| \geq 0 $$, we need $$ \cos x \geq 0 $$.

Squaring both sides (valid since both sides are non-negative when solutions exist):

$$\cos^2 x = \sin^2 x$$

$$\cos 2x = 0$$

$$2x = \frac{\pi}{2} + k\pi, \quad k \in \mathbb{Z}$$

$$x = \frac{\pi}{4} + \frac{k\pi}{2}$$

The general solutions are $$ x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}, \ldots $$

We need $$ \cos x \geq 0 $$, which holds when $$ x \in [-\frac{\pi}{2} + 2k\pi, \frac{\pi}{2} + 2k\pi] $$.

At $$ x = \frac{\pi}{4} + \frac{k\pi}{2} $$:

• $$ k = 0 $$: $$ x = \frac{\pi}{4} $$, $$ \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2} > 0 $$ ✓
• $$ k = 1 $$: $$ x = \frac{3\pi}{4} $$, $$ \cos\frac{3\pi}{4} = -\frac{\sqrt{2}}{2} < 0 $$ ✗
• $$ k = 2 $$: $$ x = \frac{5\pi}{4} $$, $$ \cos\frac{5\pi}{4} = -\frac{\sqrt{2}}{2} < 0 $$ ✗
• $$ k = 3 $$: $$ x = \frac{7\pi}{4} $$, $$ \cos\frac{7\pi}{4} = \frac{\sqrt{2}}{2} > 0 $$ ✓

So within each period of $$ 2\pi $$, we get exactly 2 valid solutions: $$ x = \frac{\pi}{4} + 2m\pi $$ and $$ x = -\frac{\pi}{4} + 2m\pi $$ for integer $$ m $$.

Solutions of the form $$ x = \frac{\pi}{4} + 2m\pi $$:

$$-4\pi \le \frac{\pi}{4} + 2m\pi \le 4\pi$$

$$-\frac{17}{8} \le m \le \frac{15}{8}$$

So $$ m = -2, -1, 0, 1 $$ — giving 4 solutions.

Solutions of the form $$ x = -\frac{\pi}{4} + 2m\pi $$:

$$-4\pi \le -\frac{\pi}{4} + 2m\pi \le 4\pi$$

$$-\frac{15}{8} \le m \le \frac{17}{8}$$

So $$ m = -1, 0, 1, 2 $$ — giving 4 solutions.

Total solutions = $$ 4 + 4 = 8 $$.

The number of solutions is 8, which corresponds to Option C.