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BITSAT 2016
Exam
BITSAT
Year
2016
Subject
Mathematics
Chapter
Trigonometric Equations

BITSAT 2016 · Mathematics · Trigonometric Equations

The general solution of the equation 2 x+2 x+2 x+1=0 is

The general solution of the equation sin⁡2x+2sin⁡x+2cos⁡x+1=0\sin 2 x+2 \sin x+2 \cos x+1=0 is

Options

  1. A3nπ−π43 \mathrm{n} \pi-\frac{\pi}{4}
  2. B2nπ+π42 \mathrm{n} \pi+\frac{\pi}{4}
  3. C2n\ppi+(−1)nsin⁡−1(13)2 \mathrm{n} \ppi+(-1)^{\mathrm{n}} \sin ^{-1} \left(\frac{1}{\sqrt{3}}\right)
  4. Dnπ−π4\mathrm{n} \pi-\frac{\pi}{4}

Correct answer

D. nπ−π4\mathrm{n} \pi-\frac{\pi}{4}

Step-by-step solution

Given, sin⁡2x+2sin⁡x+2cos⁡x+1=0⇒1+sin⁡2x+2(sin⁡x+cos⁡x)=0⇒(sin⁡x+cos⁡x)2+2(sin⁡x+cos⁡x)=0⇒(sin⁡x+cos⁡x)(sin⁡x+cos⁡x+2)=0∴sin⁡x+cos⁡x=0\sin 2 \mathrm{x}+2 \sin \mathrm{x}+2 \cos \mathrm{x}+1=0\Rightarrow 1+\sin 2 \mathrm{x}+2(\sin \mathrm{x}+\cos \mathrm{x})=0\Rightarrow(\sin x+\cos x)^{2}+2(\sin x+\cos x)=0\Rightarrow(\sin x+\cos x)(\sin x+\cos x+2)=0\therefore \sin x+\cos x=0 or sin⁡x+cos⁡x=−2\sin x+\cos x=-2 But, sin⁡x+cos⁡x=−2\sin x+\cos x=-2 is inadmissible. Since, ∣sin⁡x∣≤1,∣cos⁡x∣≤1∴sin⁡x+cos⁡x=0⇒sin⁡(x+π4)=0⇒x+π4=nπ⇒x=nπ−π4|\sin x| \leq 1,|\cos x| \leq 1\therefore \sin x+\cos x=0 \Rightarrow \sin \left(x+\frac{\pi}{4}\right)=0\Rightarrow x+\frac{\pi}{4}=n \pi \Rightarrow x=n \pi-\frac{\pi}{4}
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