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JEE Advanced 2009Paper 1
Exam
JEE Advanced
Year
2009
Paper
Paper 1
Subject
Mathematics
Chapter
Trigonometric Equations

JEE Advanced 2009 · Paper 1 · Mathematics · Trigonometric Equations

If ( ^4 x)/(2)+( ^4 x)/(3)=(1)/(5) , then

If sin⁡4x2+cos⁡4x3=15\frac{\sin ^4 x}{2}+\frac{\cos ^4 x}{3}=\frac{1}{5}, then

Options

  1. Atan⁡2x=23\tan ^2 x=\frac{2}{3}
  2. Bsin⁡8x8+cos⁡8x27=1125\frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{1}{125}
  3. Ctan⁡2x=13\tan ^2 x=\frac{1}{3}
  4. Dsin⁡8x8+cos⁡8x27=2125\frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{2}{125}

Correct answer

A. tan⁡2x=23\tan ^2 x=\frac{2}{3}

Step-by-step solution

sin⁡4x2+cos⁡4x3=15⇒sin⁡4x2+(1−sin⁡2x)23=15⇒sin⁡4x2+1+sin⁡4x−2sin⁡2x3=15⇒5sin⁡4x−4sin⁡2x+2=65⇒25sin⁡4x−20sin⁡2x+4=0⇒(5sin⁡2x−2)2=0⇒sin⁡2x=25,cos⁡2x=35,tan⁡2x=23∴sin⁡8x8+cos⁡8x27=1125\begin{aligned}& \frac{\sin ^4 x}{2}+\frac{\cos ^4 x}{3}=\frac{1}{5}\\ & \Rightarrow \quad \frac{\sin ^4 x}{2}+\frac{\left(1-\sin ^2 x\right)^2}{3}=\frac{1}{5}\\ & \Rightarrow \frac{\sin ^4 x}{2}+\frac{1+\sin ^4 x-2 \sin ^2 x}{3}=\frac{1}{5}\\ & \Rightarrow \quad 5 \sin ^4 x-4 \sin ^2 x+2=\frac{6}{5}\\ & \Rightarrow \quad 25 \sin ^4 x-20 \sin ^2 x+4=0 \\ & \Rightarrow \quad\left(5 \sin ^2 x-2\right)^2=0 \\ & \Rightarrow \sin ^2 x=\frac{2}{5}, \cos ^2 x=\frac{3}{5}, \tan ^2 x=\frac{2}{3}\\ & \therefore \quad \frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{1}{125}\\ & \end{aligned}
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