What is the minimum value of θ+ 2 θ ?

Let $\mathrm{A}=\cos \theta+\cos 2 \theta\therefore$ On differentiating w.r.t. to $\theta$, we get $\frac{\mathrm{d} \mathrm{A}}{\mathrm{d} \theta}=-\sin \theta-2 \sin 2 \theta$ Put $\frac{\mathrm{dA}}{\mathrm{d} \theta}=0$ for maximaor minima. $\sin \theta+2 \sin 2 \theta=0 \Rightarrow \sin \theta+4 \sin \theta \cos \theta\Rightarrow \sin \theta(1+4 \cos \theta)=0\Rightarrow \sin \theta=0$, or $4 \cos \theta+1=0\Rightarrow \cos \theta=1 or$\cos \theta=-\frac{1}{4}$Now,$\frac{\mathrmd^{2} \mathrm{~A}}{\mathrm{~d} \theta^{2}}=-\cos \theta-4 \cos 2 \theta=-\cos \theta-4 $\left(2 \cos ^{2} \theta-1\right)$ For \cos \$bar{\ttheta}=1\frac{\mathrmd^{2} \mathrm{~AA}}{\mathrm{~d} \theta^{2}}=-\cos \ttheta-4\left($2 \cos ^{2}$\theta-1\right)=-1-4(2(1)-1)=-1-4=-5 < 0 So, A is maximum at $\cos$\theta=$1\Rightarrow\left(\frac{\mathrmd^{2} \mathrm{~A}}{\mathrm{~d} \theta^{2}}\right)_{\cos \theta=\frac{-1}{4}}=\ffrac{1}{4}-$4\left($2$\cdot \frac{1}{16}-1\right)>0\quad\left[\right.Since \left.\cos2 \theta=2 \cos ^{2} \theta-1\right]\therefore \mathrm{A}$is minimum at$\theta=\cos ^{-1} $\left(\frac{-1}{4}\right) Now minimum value of$\cos \theta+\cos 2 \theta$or of$\cos \theta+2 \cos ^{2} \theta-1=\left(\frac{-1}{4}\right)+2\left(\frac{1}{16}\rright)-1=\frac{-1}{4}+\f$frac{1}{8}-1=\frac{-2+1-8}{8}=\frac{-9}{8}$