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JEE Advanced
Exam
JEE Advanced
Subject
Mathematics
Chapter
Trigonometry

JEE Advanced · Mathematics · Trigonometry

If α satisfies the equation 2 2 ^3 x-54 2 ^3 x=19 , then possible...

If α\alpha satisfies the equation 22tan⁡3x−542cot⁡3x=192 \sqrt{2} \tan ^3 x-54 \sqrt{2} \cot ^3 x=19, then possible value of ( 2tan⁡2α+2tan⁡α)\left.2 \tan ^2 \alpha+\sqrt{2} \tan \alpha\right) can be equal to:

Options

  1. A6
  2. B12
  3. C2
  4. D−42-4 \sqrt{2}

Correct answer

C. 2

Step-by-step solution

(2\tanx)3+(−(\sqrt{2} \tanx)^3+(-3 \sqrt{2} \cotx)^3+(-1)3=181)^3=18⇒2\tanx−32\cotx−1=0⇒2tan⁡2x−tan⁡x−32=0 and tan⁡x=1±1+2422=1±522∴ and tan⁡x=32 or −2⇒ and 2tan⁡2α+2tan⁡α=12,2\begin{array}{ll}\Rightarrow & \sqrt{2} \tanx-3 \sqrt{2} \cotx-1=0 \quad \Rightarrow \quad \sqrt{2} \tan ^2x-\tan x-3 \sqrt{2}=0 \\ \text{ and } \tan x=\frac{1 \pm \sqrt{1+24}}{2 \sqrt{2}}=\frac{1 \pm 5}{2 \sqrt{2}}\\ \therefore \text{ and } \tan x=\frac{3}{\sqrt{2}} \quad \text { or } \quad-\sqrt{2}\\ \Rightarrow \text{ and } 2 \tan ^2 \alpha+\sqrt{2} \tan \alpha=12,2\end{array}$$
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