<p>Find the number of solutions of \(\cos x = |1 + \sin x|\), \(0 < x < 3\pi\)</p>
Step-by-Step Solution
Key Concept: Convert the absolute value equation by recognizing that $1 + \sin x > 0$ always, then use the cosine difference formula to solve.
<p><strong>Step 1:</strong> Since $1 + \sin x > 0$ for all $x$, we have $\cos x = 1 + \sin x$.</p><p><strong>Step 2:</strong> Rearranging: $\cos x - \sin x = 1$</p><p><strong>Step 3:</strong> Dividing both sides by $\sqrt{a^2 + b^2} = \sqrt{2}$:</p><p>$$\frac{1}{\sqrt{2}}\cos x - \frac{1}{\sqrt{2}}\sin x = \frac{1}{\sqrt{2}}$$</p><p><strong>Step 4:</strong> This becomes:$$\cos\frac{\pi}{4}\cos x - \sin\frac{\pi}{4}\sin x = \cos\frac{\pi}{4}$$</p><p>$$\cos\left(x + \frac{\pi}{4}\right) = \cos\frac{\pi}{4}$$</p><p><strong>Step 5:</strong> Therefore: $x + \frac{\pi}{4} = 2n\pi \pm \frac{\pi}{4}$</p><p>This gives $x = 2n\pi$ or $x = 2n\pi - \frac{\pi}{2}$</p><p><strong>Step 6:</strong> For $0 < x < 3\pi$: $x = 0, \frac{3\pi}{2}, 2\pi$</p><p>∴ Number of solutions = <strong>3</strong></p>
Correct Answer: 3