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NDA 2020 Maths · Mathematics · Trigonometric Equations · medium

The smallest positive value of x satisfying √3cos x+sin x=√2 is

Single correct option

The smallest positive value of xx satisfying 3cos⁡x+sin⁡x=2\sqrt3\cos x+\sin x=\sqrt2 is
  1. A. π12\dfrac\pi{12}
  2. B. π4\dfrac\pi4
  3. C. 5π12\dfrac{5\pi}{12}
  4. D. 7π12\dfrac{7\pi}{12}
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Answer: (C) 5π12\dfrac{5\pi}{12}

Write 3cos⁡x+sin⁡x=2cos⁡(x−π6)\sqrt3\cos x+\sin x=2\cos\left(x-\frac\pi6\right). Setting it equal to 2\sqrt2 gives cos⁡(x−π6)=12\cos\left(x-\frac\pi6\right)=\frac1{\sqrt2}, so x−π6=±π4+2kπx-\frac\pi6=\pm\frac\pi4+2k\pi. The value x=π6−π4x=\frac\pi6-\frac\pi4 is negative, so the smallest positive solution is x=5π12x=\frac{5\pi}{12}.
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Original question written in the NDA 2020 Maths pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake