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

Correct Answer is : {2π3,4π3}\{2π/3, 4π/3\}{32π,34π}

Correct Answer is : {2π3,4π3}\{2π/3, 4π/3\}{32π,34π}

Test Index

EAMCET Engineering 2003 Solved Paper

Section: Mathematics

Question : 159 of 160

Marks: +1, -0

The solution set of

(5+4\cos\theta)(2\cos\theta+1)=0

[0,2\pi]

Solution:

We have,

(5+4\cos\theta)(2\cos\theta+1)=0

. . . (i)

\cos\theta = \frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}

\therefore \cos\theta = \frac{1-t^2}{1+t^2} \quad \left[\text{put } \tan\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 [5+5t^2+4-4t^2][2-2t^2+1+t^2]=0

\Rightarrow (t^2+9)(3-t^2)=0

\therefore t = \pm \sqrt{3}

\Rightarrow \tan\frac{\theta}{2} = \sqrt{3} \text{ or } \tan\frac{\theta}{2} = -\sqrt{3}

\Rightarrow \frac{\theta}{2} = \frac{\pi}{3} \text{ or } \frac{\theta}{2} = \frac{2\pi}{3}

\therefore \theta = \frac{2\pi}{3} \text{ or } \frac{4\pi}{3}

\;

Go to Question:

Prev Question

Next Question