Differentiation (Part-38) | S N De

 

Solve :

$\,\,1(i)\,\, \tan\theta +\cot2\theta =2 \\ \Rightarrow \frac{\sin\theta }{\cos\theta }+\frac{\cos2\theta }{\sin2\theta }=2 \\ \Rightarrow \frac{\sin\theta \sin2\theta +\cos\theta \cos2\theta }{\cos\theta \sin2\theta }=2 \\ \Rightarrow \frac{\cos2\theta \cos\theta +\sin\theta \sin2\theta}{\cos\theta \sin2\theta }=2 \\ \Rightarrow \frac{\cos\theta }{\cos\theta \sin2\theta }=2\\ \Rightarrow \frac{1}{\sin2\theta }=2\,\,[\text{since},\,\,\cos\theta \neq 0]\\ \therefore \sin2\theta =\frac 12=\sin\frac{\pi}{6}\\ \Rightarrow 2\theta =n\pi+(-1)^n\pi/6,\,\,n \in\mathbb Z\\ \Rightarrow \theta =\frac{n\pi}{2}+(-1)^n\frac{\pi}{12}\,\,\text{(ans.)}$

$\,\,1(ii)\,\,\cot\theta +\tan\theta =2\sec\theta \quad (0^{\circ}<\theta <360^{\circ})\\ \Rightarrow \frac{\cos\theta }{\sin\theta }+\frac{\sin\theta }{\cos\theta }=2\sec\theta \\ \Rightarrow \frac{\cos^2\theta +\sin^2\theta }{\sin\theta \cos\theta }=2\sec\theta  \\ \Rightarrow \frac{1}{\sin\theta \cos\theta }=\frac{2}{\cos\theta } \\ \Rightarrow \frac{1}{\sin\theta} =2 \\ \Rightarrow \sin\theta =\frac 12 \\ \Rightarrow \theta =30^{\circ}, 150^{\circ}\quad (0^{\circ}<\theta <360^{\circ})$

$\,\,1(iii)\,\,\cot2x=\cos x+\sin x \\ \Rightarrow \frac{\cos2x}{\sin2x}=\cos x+\sin x \\ \Rightarrow \cos2x=(\cos x+\sin x)\sin2x \\ \Rightarrow \cos^2x-\sin^2x=(\cos x+\sin x)\sin2x \\ \Rightarrow (\cos x+\sin x)(\cos x-\sin x)\\-(\cos x+\sin x)\sin2x=0 \\ \Rightarrow (\cos x+\sin x)(\cos x-\sin x-\sin2x)=0 \\ \Rightarrow \cos x+\sin x=0\rightarrow(1), \\ \text{or,}\,\,\cos x-\sin x-\sin2x=0\rightarrow(2) $

From $\,(1)\,$, we get, $\,\, \sin x=-\cos x \\ \Rightarrow \tan x=-1=\tan\left(-\frac{\pi}{4}\right) \\ \Rightarrow x=n\pi-\frac{\pi}{4}\,,\,\,\,n\in\mathbb Z \rightarrow(2a)$

Again, by $\,(2)\,$, we get, $\,\,\,\cos x-\sin x=\sin2x \\ \Rightarrow(\cos x-\sin x)^2=\sin^22x \\ \Rightarrow1-\sin2x=\sin^22x\\ \Rightarrow\sin^22x+\sin2x-1=0\\ \Rightarrow\sin2x=\frac{-1\pm\sqrt5}{2}\rightarrow(3)$

From $\,(3)\,\,$, we get, $\,\sin2x=\frac{-1+\sqrt5}{2}=\sin\alpha,\,\,\text{say} \\ \Rightarrow2x=n\pi+(-1)^n\alpha,\,\,n\in\mathbb Z \\ \Rightarrow x=\frac{n\pi}{2}+(-1)^n\frac{\alpha}{2}\rightarrow(4)$

Also, from $\,(3)\,$ we note that, $\,\,\sin2x \neq \frac{-1-\sqrt5}{2}\,\,\text{as}\,\,-1\leq \sin2x\leq 1$

Hence, the general solution is given by $\,(2a),\,\,(4).$

$\,1(iv)\,\,\sin x+\cos x=\sqrt2\cos2x \\ \Rightarrow \sin x+\cos x=\sqrt2(\cos^2x-\sin^2x)\\ \Rightarrow \sin x+\cos x=\sqrt2(\cos x+\sin x)\\ \times(\cos x-\sin x)\\ \Rightarrow (\cos x+\sin x)(\sqrt2 \cos x \\-\sqrt2\sin x-1)=0 \\ \therefore \cos x+\sin x=0\rightarrow(1), \\ \text{or,}\,\,\sqrt2\cos x-\sqrt2\sin x-1=0 \rightarrow(2).$

From $\,(1)\,$, we get, $\,\,\sin x=-\cos x \\ \Rightarrow \tan x=-1=\tan\left(-\pi/4\right) \\ \Rightarrow x= n\pi-\frac{\pi}{4},\,\,n\in\mathbb Z.\rightarrow(3)$

By $\,(2)\,\,$ we get, $\,\,\frac{1}{\sqrt2}\cos x-\frac{1}{\sqrt2}\sin x=\frac 12 \\ \Rightarrow \cos(x+\frac{\pi}{4})=\cos\left(\pm\frac{\pi}{3}\right) \\ \Rightarrow  x+\frac{\pi}{4}=2n\pi \pm \frac{\pi}{3} \\ \Rightarrow x=2n\pi+\frac{\pi}{12},\,\,2n\pi-\frac{7\pi}{12},\,\,n\in\mathbb  Z.\rightarrow(4)$

Hence, the general solution is given by $\,(3),\,\,(4).$

$\,1(v)\,\, \cot x-\cot 2x=2 \\ \Rightarrow \frac{\cos x}{\sin x}-\frac{\cos2x}{\sin2x}=2  \\ \Rightarrow \frac{\cos x\sin2x-\sin x\cos 2x}{\sin x\sin2x}=2  \\ \Rightarrow  \frac{\sin(2x-x)}{\sin2x\sin x}=2 \\ \Rightarrow \frac{\sin x}{\sin2x\sin x}=2  \\ \Rightarrow \frac{1}{\sin 2x}=2 \,\,[\text{since,}\,\,\sin x\neq 0] \\ \Rightarrow  \sin 2x=\frac 12=\sin\frac{\pi}{6} \\ \Rightarrow 2x=n\pi+(-1)^n\frac{\pi}{6}  \\ \Rightarrow x=\frac{n\pi}{2}+(-1)^n\frac{\pi}{12},\,\,\, n\in\mathbb Z.$

$\,1(vi)\,\,\cos x+\cos2x+\cos3x=0 \\ \Rightarrow (\cos x+\cos3x)+\cos2x=0 \\ \Rightarrow  2\cos\left(\frac{x+3x}{2}\right)\cos\left(\frac{3x-x}{2}\right)+\cos2x=0 \\ \Rightarrow 2\cos2x\cos x+\cos2x=0 \\ \Rightarrow \cos2x(2\cos x+1)=0 \\ \therefore \cos2x=0 \rightarrow(1), \\ \text{or},\,\,2\cos x+1=0\rightarrow(2).$

From $\,(1)\,\,$ we get, $\,\,\,2x=(2n+1)\frac{\pi}{2} \\ \Rightarrow  x=(2n+1)\frac{\pi}{4},\,\,\, n \in\mathbb Z \rightarrow(3)$

By $\,\,(2)\,\,$ we get, $\quad\cos x=-\frac 12=\cos\frac{2\pi}{3} \\ \Rightarrow x=2n\pi \pm \frac{2\pi}{3},\,\,n \in\mathbb Z \rightarrow(4)$

Hence, from $\,\,(3),\,(4)\,\,$ we get the required general solution.