Trigonometry & Inverse Trigonometry Question - None

Question

If sin x + cos x = 1 + sin x cos x, then x is

Accepted Answer

Step 1: Let \sin x + \cos x = t

Step 2: Then t^2 = 1 + 2\sin x \cos x, so \sin x \cos x = \frac{t^2-1}{2}

Step 3: Substituting into the given equation: t = 1 + \frac{t^2-1}{2}

Step 4: 2t = 2 + t^2 - 1 \Rightarrow t^2 - 2t + 1 = 0 \Rightarrow (t-1)^2 = 0 \Rightarrow t = 1

Step 5: So \sin x + \cos x = 1

Step 6: \sqrt{2}\sin\left(x + \frac{\pi}{4}\right) = 1 \Rightarrow \sin\left(x + \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}

Step 7: x + \frac{\pi}{4} = n\pi + (-1)^n\frac{\pi}{4}

Step 8: For even n: x = 2n\pi; for odd n: x = n\pi + \frac{\pi}{4}

∴ Answer is B.

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