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NDA 2022
Exam
NDA
Year
2022
Subject
Mathematics
Chapter
Trigonometric Equations

NDA 2022 · Mathematics · Trigonometric Equations

If 1+2( x+ x)( x- x)=0 where 0 < x < 360^ ^ , then how many values...

If 1+2(sin⁡x+cos⁡x)(sin⁡x−cos⁡x)=01+2(\sin x+\cos x)(\sin x-\cos x)=0 where 0<x<360∘0 < x < 360^{{}^{\circ}}, then how many values does xx take?

Options

  1. AOnly one value
  2. BOnly two values
  3. COnly three values
  4. DFour values

Correct answer

D. Four values

Step-by-step solution

1+2(sin⁡x+cos⁡x)(sin⁡x−cos⁡x)=0⇒1+2(sin⁡2x−cos⁡2x)=0⇒1−2cos⁡22x=0\begin{aligned} & 1+2(\sin x+\cos x)(\sin x-\cos x)=0 \\ & \Rightarrow 1+2\left(\sin ^2 x-\cos ^2 x\right)=0 \\ & \Rightarrow 1-2 \cos ^2 2 x=0\end{aligned}⇒cos⁡22x=12⇒cos⁡2x=12⇒x=nπ±π8∴x=0,π−π8π+π8,2π−π8. Four values. \begin{aligned} & \Rightarrow \cos ^2 2 x=\frac{1}{2} \quad \Rightarrow \cos 2 x=\frac{1}{\sqrt{2}} \\ & \Rightarrow x=n \pi \pm \frac{\pi}{8} \quad \therefore \quad x=0, \pi-\frac{\pi}{8} \\ & \pi+\frac{\pi}{8}, 2 \pi-\frac{\pi}{8}. \text{ Four values. }\end{aligned}$
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