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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

JEE Main · Mathematics · Trigonometric Ratios & Identities

The solution set of (5+4 θ)(2 θ+1)=0 in the interval [0,2 π] , is :

The solution set of (5+4cos⁡θ)(2cos⁡θ+1)=0(5+4 \cos \theta)(2 \cos \theta+1)=0 in the interval [0,2π][0,2 \pi], is :

Options

  1. A\left\{\frac{\pi}{3}, \frac{2 \pi}{3}\right\}
  2. B\left\{\frac{\pi}{3}, \pi\right\}
  3. C\left\{\frac{2 \pi}{3}, \frac{4 \pi}{3}\right\}
  4. D\left\{\frac{2 \pi}{3}, \frac{5 \pi}{3}\right\}

Correct answer

C. \left\{\frac{2 \pi}{3}, \frac{4 \pi}{3}\right\}

Step-by-step solution

We have, (5+4cos⁡θ)(2cos⁡θ+1)=0(5+4 \cos \theta)(2 \cos \theta+1)=0...(i) cos⁡θ=1−tan⁡2θ21+tan⁡2θ2∴cos⁡θ=1−t21+t2\cos \theta=\frac{1-\tan ^2 \frac{\theta}{2}}{1+\tan ^2 \frac{\theta}{2}}\therefore \quad \cos \theta=\frac{1-t^2}{1+t^2} \quad\left[\right.putput\left.\tan θ2=\frac{\theta}{2}=t\right]Then, Eq. (i) becomes \left[5+4\left(\frac{1-t^2}{1+t^2}\right)\right]\left[2\left(\frac{1-t^2}{1+t^2}\right)+1\right]=0\Rightarrow\left[5+5 t^2+4-4 t^2\right]\left[2-2 t^2+1+t^2\right]=0\Rightarrow \quad\left(t^2+9\right)\left(3-t^2\right)=0\therefore \quad t= \pm\sqrt{3}\Rightarrow \quad \tan \frac{\theta}{2}=\sqrt{3} or tan⁡θ2=−3⇒θ2=π3orθ2=2π3∴θ=2π3or4π3\tan \frac{\theta}{2}=-\sqrt{3}\Rightarrow \quad \frac{\theta}{2}=\frac{\pi}{3} or \frac{\theta}{2}=\frac{2 \pi}{3}\therefore \quad \theta=\frac{2 \pi}{3} or \frac{4 \pi}{3}
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