Question 14

Mathematics Trigonometric Equations Hard

The solution of sin x + 1 = cos x, such that 0 ≤ x < 2π, is:

(A) 0, π
(B) <p><span class="math-tex">\(0, \dfrac{\pi}{2}\)</span></p>
(C) <span class="math-tex">\(\dfrac{\pi}{2}, \dfrac{3\pi}{2}\)</span>
(D) <span class="math-tex">\(0, \dfrac{3\pi}{2}\)</span>
✓ Correct Answer: Option D Verified Solution

📘 Step-by-Step Solution & Detailed Explanation:

\[ \sin x + 1 = \cos x \] \[ \cos x - \sin x = 1 \] \[ \frac{1}{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x = \frac{1}{\sqrt{2}} \] \[ \cos \left(x + \frac{\pi}{4}\right) = \cos \frac{\pi}{4} \] \[ \left(x + \frac{\pi}{4}\right) = 2n\pi \pm \frac{\pi}{4} \] Putting \( n = 0 \) and \( 1 \), \( x = 0 \) and \( \frac{3\pi}{2} \).

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