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If \(\sin x+\cos x=\sqrt{2}\), what is (x) in \(0\leq x<2\pi\)?यदि \(\sin x+\cos x=\sqrt{2}\), तो \(0\leq x<2\pi\) में (x) का मान क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: \( \frac{\pi}{4} \)\( \frac{\pi}{4} \)
We use \(\sin x+\cos x=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)\). The maximum \(\sqrt{2}\) occurs at \(x=\frac{\pi}{4}\).\(\sin x+\cos x=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)\) है। अधिकतम \(\sqrt{2}\) तब मिलता है जब \(x=\frac{\pi}{4}\)।
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