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JEE Main 2006
Exam
JEE Main
Year
2006
Subject
Mathematics
Chapter
Trigonometric Equations

JEE Main 2006 · Mathematics · Trigonometric Equations

If 0 < x < π and x+ x=(1)/(2) , then x is

If 0<x<π0 < x < \pi and cos⁡x+sin⁡x=12\cos x+\sin x=\frac{1}{2}, then tan⁡x\tan x is

Options

  1. A(1−7)4\frac{(1-\sqrt{7})}{4}
  2. B(4−7)3\frac{(4-\sqrt{7})}{3}
  3. C−(4+7)3-\frac{(4+\sqrt{7})}{3}
  4. D(1+7)4\frac{(1+\sqrt{7})}{4}

Correct answer

C. −(4+7)3-\frac{(4+\sqrt{7})}{3}

Step-by-step solution

cos⁡x+sin⁡x=12⇒1+sin⁡2x=14⇒sin⁡2x=−34\cos x+\sin x=\frac{1}{2} \Rightarrow 1+\sin 2 x=\frac{1}{4} \Rightarrow \sin 2 x=-\frac{3}{4}, so xx is obtuse and 2tan⁡x1+tan⁡2x=−34⇒3tan⁡2x+8tan⁡x+3=0\frac{2 \tan x}{1+\tan ^2 x}=-\frac{3}{4} \Rightarrow 3 \tan ^2 x+8 \tan x+3=0 ∴tan⁡x=−8±64−366=−4±73\therefore \tan x=\frac{-8 \pm \sqrt{64-36}}{6}=\frac{-4 \pm \sqrt{7}}{3} ∵tan⁡x<0∴tan⁡x=−4−73\because \tan x < 0 \quad \therefore \tan x=\frac{-4-\sqrt{7}}{3}
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