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The solution set of $(5+4 \cos \theta)(2 \cos \theta+1)=0$ in the interval $[0,2 \pi]$, is :
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Verified Answer
The correct answer is:
$\left\{\frac{2 \pi}{3}, \frac{4 \pi}{3}\right\}$
We have,
$(5+4 \cos \theta)(2 \cos \theta+1)=0$...(i)
$\cos \theta=\frac{1-\tan ^2 \frac{\theta}{2}}{1+\tan ^2 \frac{\theta}{2}}$
$\therefore \quad \cos \theta=\frac{1-t^2}{1+t^2} \quad\left[\right.$ put $\left.\tan \frac{\theta}{2}=t\right]$
Then, Eq. (i) becomes
$\left[5+4\left(\frac{1-t^2}{1+t^2}\right)\right]\left[2\left(\frac{1-t^2}{1+t^2}\right)+1\right]=0$
$\Rightarrow\left[5+5 t^2+4-4 t^2\right]\left[2-2 t^2+1+t^2\right]=0$
$\Rightarrow \quad\left(t^2+9\right)\left(3-t^2\right)=0$
$\therefore \quad t= \pm \sqrt{3}$
$\Rightarrow \quad \tan \frac{\theta}{2}=\sqrt{3}$ or $\tan \frac{\theta}{2}=-\sqrt{3}$
$\Rightarrow \quad \frac{\theta}{2}=\frac{\pi}{3}$ or $\frac{\theta}{2}=\frac{2 \pi}{3}$
$\therefore \quad \theta=\frac{2 \pi}{3}$ or $\frac{4 \pi}{3}$
$(5+4 \cos \theta)(2 \cos \theta+1)=0$...(i)
$\cos \theta=\frac{1-\tan ^2 \frac{\theta}{2}}{1+\tan ^2 \frac{\theta}{2}}$
$\therefore \quad \cos \theta=\frac{1-t^2}{1+t^2} \quad\left[\right.$ put $\left.\tan \frac{\theta}{2}=t\right]$
Then, Eq. (i) becomes
$\left[5+4\left(\frac{1-t^2}{1+t^2}\right)\right]\left[2\left(\frac{1-t^2}{1+t^2}\right)+1\right]=0$
$\Rightarrow\left[5+5 t^2+4-4 t^2\right]\left[2-2 t^2+1+t^2\right]=0$
$\Rightarrow \quad\left(t^2+9\right)\left(3-t^2\right)=0$
$\therefore \quad t= \pm \sqrt{3}$
$\Rightarrow \quad \tan \frac{\theta}{2}=\sqrt{3}$ or $\tan \frac{\theta}{2}=-\sqrt{3}$
$\Rightarrow \quad \frac{\theta}{2}=\frac{\pi}{3}$ or $\frac{\theta}{2}=\frac{2 \pi}{3}$
$\therefore \quad \theta=\frac{2 \pi}{3}$ or $\frac{4 \pi}{3}$
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