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Harvard-MIT Math Tournament

Algebra Difficulty 4.3 AIME Prove it United States

Problem:
Find all values of $x$ with $0 \leq x < 2\pi$ that satisfy $\sin x + \cos x = \sqrt{2}$ .

Solution

Solution:
Squaring both sides gives $\sin^2 x + \cos^2 x + 2 \sin x \cos x = 1 + \sin 2x = 2$ , so $x = \frac{\pi}{4}, \frac{5\pi}{4}$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.