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

Question

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

Aicharya · Accepted Answer

We have, $(5+4 \cos 	heta)(2 \cos 	heta+1)=0$...(i) $\cos 	heta=\frac{1-	an ^2 \frac{	heta}{2}}{1+	an ^2 \frac{	heta}{2}}$ $	herefore \quad \cos 	heta=\frac{1-t^2}{1+t^2} \quad\left[ight.$ put $\left.	an \frac{	heta}{2}=tight]$ Then, Eq. (i) becomes $\left[5+4\left(\frac{1-t^2}{1+t^2}ight)ight]\left[2\left(\frac{1-t^2}{1+t^2}ight)+1ight]=0$ $\Rightarrow\left[5+5 t^2+4-4 t^2ight]\left[2-2 t^2+1+t^2ight]=0$ $\Rightarrow \quad\left(t^2+9ight)\left(3-t^2ight)=0$ $	herefore \quad t= \pm \sqrt{3}$ $\Rightarrow \quad 	an \frac{	heta}{2}=\sqrt{3}$ or $	an \frac{	heta}{2}=-\sqrt{3}$ $\Rightarrow \quad \frac{	heta}{2}=\frac{\pi}{3}$ or $\frac{	heta}{2}=\frac{2 \pi}{3}$ $	herefore \quad 	heta=\frac{2 \pi}{3}$ or $\frac{4 \pi}{3}$

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

Solution