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JEE Main 201411 AprilOnline
Exam
JEE Main
Year
2014
Date
11 April
Mode
Online
Subject
Mathematics
Chapter
Trigonometric Equations

JEE Main 2014 · 11 April · Online · Mathematics · Trigonometric Equations

If 2 θ+ θ=1(θ ≠ (π)/(2)) , then 7 θ+6 θ is equal to:

If 2cos⁡θ+sin⁡θ=1(θ≠π2)2 \cos \theta+\sin \theta=1\left(\theta \neq \frac{\pi}{2}\right), then 7cos⁡θ+6sin⁡θ7 \cos \theta+6 \sin \theta is equal to:

Options

  1. A12\frac{1}{2}
  2. B2
  3. C112\frac{11}{2}
  4. D465\frac{46}{5}

Correct answer

D. 465\frac{46}{5}

Step-by-step solution

Given 2cos⁡θ+sin⁡θ=12 \cos \theta+\sin \theta=1 Squaring both sides, we get (2cos⁡θ+sin⁡θ)2=12⇒4cos⁡2θ+sin⁡2θ+4sin⁡θcos⁡θ=1⇒3cos⁡2θ+(cos⁡2θ+sin⁡2θ)+4sin⁡θcos⁡θ=1⇒3cos⁡2θ+λ̸+4sin⁡θcos⁡θ=1⇒3cos⁡2θ+4sin⁡θcos⁡θ=0⇒cos⁡θ(3cos⁡θ+4sin⁡θ)=0⇒3cos⁡θ+4sin⁡θ=0⇒3cos⁡θ=−4sin⁡θ⇒−34=tan⁡θ=sec⁡2θ−1=−34(∵tan⁡2θ=sec⁡2θ−1)⇒sec⁡2θ−1=(−34)2=916⇒sec⁡2θ=916+1=2516⇒sec⁡θ=54\begin{aligned}&(2 \cos \theta+\sin \theta)^2=1^2 \\ &\Rightarrow 4 \cos ^2 \theta+\sin ^2 \theta+4 \sin \theta \cos \theta=1 \\ &\Rightarrow 3 \cos ^2 \theta+\left(\cos ^2 \theta+\sin ^2 \theta\right)+4 \sin \theta \cos \theta\\ &=1 \\ &\Rightarrow 3 \cos ^2 \theta+\not \lambda+4 \sin \theta \cos \theta=1 \\ &\Rightarrow 3 \cos ^2 \theta+4 \sin \theta \cos \theta=0 \\ &\Rightarrow \cos \theta(3 \cos \theta+4 \sin \theta)=0 \\ &\Rightarrow 3 \cos \theta+4 \sin \theta=0 \\ &\Rightarrow 3 \cos \theta=-4 \sin \theta\\ &\Rightarrow \frac{-3}{4}=\tan \theta=\sqrt{\sec ^2 \theta-1}=\frac{-3}{4}\\ &\left(\because \tan ^2 \theta=\sqrt{\sec ^2 \theta-1}\right)\\ &\Rightarrow \sec ^2 \theta-1=\left(\frac{-3}{4}\right)^2=\frac{9}{16}\\ &\Rightarrow \sec 2 \theta=\frac{9}{16}+1=\frac{25}{16} \Rightarrow \sec \theta=\frac{5}{4} \end{aligned}or cos⁡θ=45\cos \theta=\frac{4}{5} Now, sin⁡2θ+cos⁡2θ=1\sin ^2 \theta+\cos ^2 \theta=1⇒sin⁡2θ+(45)2=1sin⁡2θ+45=1⇒sin⁡2θ=1−1625=925sin⁡θ=±35\begin{aligned}&\Rightarrow \sin ^2 \theta+\left(\frac{4}{5}\right)^2=1 \\ &\sin ^2 \theta+\frac{4}{5}= 1 \Rightarrow \sin ^2 \theta=1-\frac{16}{25}=\frac{9}{25} \\ &\sin \theta=\pm \frac{3}{5} \end{aligned}$TakingTaking\quad \left(\sin \theta=+\frac{3}{5}\right)because(sin⁡θ=−35)\quad because \left(\sin \theta=-\frac{3}{5}\right) cannot satisfy the given equation. Therefore; 7 cos⁡θ+6sin⁡\cos \theta+6 \sin\theta =7 \times 45+6×35=285+185=465\frac{4}{5}+6 \times \frac{3}{5}=\frac{28}{5}+\frac{18}{5}=\frac{46}{5}
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