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JEE Advanced
Exam
JEE Advanced
Subject
Mathematics
Chapter
Trigonometry

JEE Advanced · Mathematics · Trigonometry

Set of values of x of the equation 2 x= ( ^3 x- ^3 x) is -

Set of values of xx of the equation cos⁡2x=(cos⁡3x−sin⁡3x)is−\cos 2 x= \left(\cos ^3 x-\sin ^3 x\right) is -

Options

  1. A\left\{\mathrm{x} \left\lvert\, x=n\mathrm{x}=\mathrm{n} \pi+\frac{\pi}{4}\right., \mathrm{n} \in I\mathrm{I}\right\}
  2. B{x∣x=2nπ,n∈\{\mathrm{x} \mid \mathrm{x}=2 \mathrm{n} \pi, \mathrm{n} \in\mathrm{I}}\}
  3. C\left\{\mathrm{x} \left\lvert\, \mathrm{x}=2 \mathrm{n} \pi+\frac{\pi}{2}\right., \mathrm{n} \in I\mathrm{I}\right\}
  4. D\left\{\mathrm{x} \left\lvert\, x=n\mathrm{x}=\mathrm{n} \pi-\frac{\pi}{4}\right., \mathrm{n} \in I\mathrm{I}\right\}

Correct answer

C. \left\{\mathrm{x} \left\lvert\, \mathrm{x}=2 \mathrm{n} \pi+\frac{\pi}{2}\right., \mathrm{n} \in I\mathrm{I}\right\}

Step-by-step solution

cos⁡2x−sin⁡2x=(cos⁡x−sin⁡x)(1+sin⁡xcos⁡x)cos⁡x−sin⁡x=0 or cos⁡x+sin⁡x=1+sin⁡xcos⁡xcos⁡x=sin⁡x or (cos⁡x+sin⁡x)=1+(cos⁡x+sin⁡x)2−12tan⁡x=1 Let cos⁡x+sin⁡x=tx=nπ+π4,n∈It=1+t2−12t2−2t+1=0t=1sin⁡x+cos⁡x=1cos⁡x⋅12+sin⁡x⋅12=12cos⁡(x−π4)=cos⁡π4x−π4=2nπ±π4X=2nπ+π2x=2nπ\begin{aligned} & \cos ^2 x-\sin ^2 x=(\cos x-\sin x)(1+\sin x \cos x) \\ & \cos x-\sin x=0 \text { or } \cos x+\sin x=1+\sin x \cos x \\ & \cos x=\sin x \\ & \text { or }(\cos x+\sin x)=1+\frac{(\cos x+\sin x)^2-1}{2} \\ & \tan \mathrm{x}=1 \\ & \text { Let } \cos x+\sin x=t \\ & \mathrm{x}=\mathrm{n} \pi+\frac{\pi}{4}, \mathrm{n} \in \mathrm{I} \quad \mathrm{t}=1+\frac{\mathrm{t}^2-1}{2} \\ & \mathrm{t}^2-2 \mathrm{t}+1=0 \\ & \mathrm{t}=1 \\ & \sin x+\cos x=1 \\ & \cos x \cdot \frac{1}{\sqrt{2}}+\sin x \cdot \frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}} \\ & \cos \left(x-\frac{\pi}{4}\right)=\cos \frac{\pi}{4} \\ & \mathrm{x}-\frac{\pi}{4}=2 \mathrm{n} \pi \pm \frac{\pi}{4} \\ & \mathrm{X}=2 \mathrm{n} \pi+\frac{\pi}{2} \\ & x=2 n \pi \end{aligned}
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