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Question 149 of 177

Q.The principal solutions of cot⁡x=−3\cot x = -\sqrt{3} are ________.

(a) π6,5π6\dfrac{\pi}{6}, \dfrac{5\pi}{6}
(b) 5π6,7π6\dfrac{5\pi}{6}, \dfrac{7\pi}{6}
(c) 5π6,11π6\dfrac{5\pi}{6}, \dfrac{11\pi}{6}
(d) π6,11π6\dfrac{\pi}{6}, \dfrac{11\pi}{6}
Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2019MCQ· 1mImportance★★★★★
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cot⁡x=−3  ⟺  tan⁡x=−13\cot x=-\sqrt3 \iff \tan x = -\dfrac{1}{\sqrt3}; find the principal solutions in [0,2π)[0,2\pi).

cot⁡x=−3  ⟹  tan⁡x=−13\cot x = -\sqrt3 \implies \tan x = -\dfrac{1}{\sqrt3}

Reference angle: tan⁡π6=13\tan\dfrac\pi6 = \dfrac{1}{\sqrt3}. Since tan⁡x\tan x is negative, xx lies in the 2nd or 4th quadrant.

2nd quadrant: x=π−π6=5π6x = \pi - \dfrac\pi6 = \dfrac{5\pi}{6}

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