Differentiation (Part-38) | S N De

 $\,1.\,$ Solve :

$\,(i)\,\,4\sin\theta\cos\theta=1+2\cos\theta-2\sin\theta \\ [0<\theta<\pi]$

Sol. $\,\,4\sin\theta\cos\theta=1+2\cos\theta-2\sin\theta \\ \Rightarrow 4\sin\theta\cos\theta+2\sin\theta-2\cos\theta-1=0 \\ \Rightarrow 2\sin\theta(2\cos\theta+1)-1(2\cos\theta+1)=0 \\ \Rightarrow (2\cos\theta+1)(2\sin\theta-1)=0 \\ \therefore 2\cos\theta+1=0\rightarrow(1),\\2\sin\theta-1=0\rightarrow(2)$

Now, from (1) we get,  $\cos\theta=-\frac 12 \\ \Rightarrow \theta=\frac{2\pi}{3}\,\,[\text{Since,}\,\,0<\theta<\pi]$

From (2), we get , $\,\,\sin\theta=\frac 12 \\ \Rightarrow \theta=\frac{\pi}{6},\,\frac{5\pi}{6}.\,\,[\text{Since,}\,\,0<\theta<\pi]$

$\,\,(ii)\,1+2\sin\theta\cos\theta-2\sin\theta-\cos\theta=0\\ [0^{\circ}\leq \theta \leq 360^{\circ}]$

Sol. $\,\,1+2\sin\theta\cos\theta-2\sin\theta-\cos\theta=0\\ \Rightarrow 1-\cos\theta-2\sin\theta(1-\cos\theta)=0 \\ \Rightarrow (1-\cos\theta)-2\sin\theta(1-\cos\theta)=0 \\ \Rightarrow (1-\cos\theta)(1-2\sin\theta)=0 \\ \therefore 1-\cos\theta=0 \rightarrow(1) \\ \text{and}\,\,1-2\sin\theta=0\rightarrow(2)$

Hence, from $\,(1)\,$ we get, $\,\,\cos\theta=1 \\ \Rightarrow \theta=0, 2\pi$

and from $\,(2)\,$ we get, $\,\,\sin\theta=\frac 12 \\ \Rightarrow \theta=\frac{\pi}{6},\,\,\frac{5\pi}{6}.$

$\,(iii)\,\,1-2\sin\theta-2\cos\theta+\cot\theta=0\,\,[0<\theta<2\pi] \\ \Rightarrow 1-2\sin\theta-2\cos\theta+\frac{\cos\theta}{\sin\theta}=0 \\ \Rightarrow \sin\theta-2\sin^2\theta-2\sin\theta\cos\theta+\cos\theta=0 \\ \Rightarrow \sin\theta(1-2\sin\theta)+\cos\theta(1-2\sin\theta)=0 \\ \Rightarrow (1-2\sin\theta)(\sin\theta+\cos\theta)=0 \\ \therefore 1-2\sin\theta=0 \rightarrow(1), \\ \sin\theta+\cos\theta=0 \rightarrow(2)$

Hence, from $\,(1)\,$, we get 

$\,\,\sin\theta=\frac 12 \Rightarrow \theta=\frac{\pi}{6}, \frac{5\pi}{6}.\,\,[0<\theta<2\pi]$

Again, from $\,(2)\,$, 

we get $\,\,\sin\theta=-\cos\theta \\ \Rightarrow \tan\theta=-1 \\ \Rightarrow \theta=\frac{3\pi}{4}, \frac{7\pi}{4}\,\,[0<\theta<2\pi]$

$\,(iv).\,\tan x+\tan 2x+\tan3x=0 \\ \Rightarrow \tan x+\tan2x+\tan(x+2x)=0 \\ \Rightarrow \tan x+\tan2x+\frac{\tan x+\tan 2x}{1-\tan x\tan2x}=0 \\ \Rightarrow (\tan x+\tan2x)\left(1+\frac{1}{1-\tan x\tan2x}\right)=0 \\ \Rightarrow (\tan x+\tan2x)(2-\tan x\tan2x)=0 \\ \Rightarrow(\tan x+\tan2x)\left(2-\tan x.\frac{2\tan x}{1-\tan^2x}\right)=0 \\ \Rightarrow (\tan x+\tan2x) \\ \times \left[2(1-\tan^2x)-2\tan^2x\right]=0 \\ \Rightarrow(\tan x+\tan2x)(2-4\tan^2x)=0 \\ \therefore (\tan x+\tan2x)=0\rightarrow(1), \\ \text{and}\,\,(2-4\tan^2x)=0\rightarrow(2)$

From $\,\,(1)\,\,$ we get, $\,\,\tan x+\tan 2x=0 \\ \Rightarrow \tan 2x=-\tan x \\ \Rightarrow 2x=n\pi-x\,\,[\text{where n=any integer}]\\ \Rightarrow 2x+x=n\pi \\ \Rightarrow 3x=n\pi \\ \Rightarrow x=\frac{n\pi}{3} $ 

From $\,\,(2)\,\,$ we get, $\,\,\tan^2x=\frac 12 \\ \Rightarrow \tan x=\pm\frac{1}{\sqrt2}\\ \Rightarrow \tan x=\tan(\pm \alpha)\\ \Rightarrow x=n\pi\pm\alpha,\,\,\text{where}\,\,\tan\alpha=\frac{1}{\sqrt2}$

Hence, the required general solution of the given euation is given by : 

$\,\,x=\frac{n\pi}{3},\,\,n\pi\pm\alpha,\\\text{where},\,\,\tan\alpha=\frac{1}{\sqrt2},\,\,n\,\text{being any integer.}$

$\,(v)\,\,\,\tan x+\tan2x+\tan3x\\=\tan x\tan2x\tan3x \\ \Rightarrow \tan x+\tan2x\\=-\tan3x+\tan x\tan2x\tan3x \\ \Rightarrow \tan x+\tan2x\\=-\tan3x(1-\tan x\tan2x) \\ \Rightarrow \frac{\tan x+\tan 2x}{1-\tan x\tan2x}=-\tan3x \\ \Rightarrow \tan(x+2x)=-\tan3x \\ \Rightarrow 2\tan3x=0 \\ \Rightarrow \tan3x=0\\ \Rightarrow 3x=n\pi,(\,\,n\,\text{being any integer}) \\ \Rightarrow x=\frac{n\pi}{3}$

$\,(vi)\,\,\tan3\theta+\tan\theta=2\tan2\theta\\ \Rightarrow \frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}+\tan\theta=2.\frac{2\tan\theta}{1-\tan^2\theta} \\ \Rightarrow \frac{3\tan\theta-\tan^3\theta+\tan\theta-3\tan^3\theta}{1-3\tan^2\theta}=\frac{4\tan\theta}{1-\tan^2\theta} \\ \Rightarrow \frac{4\tan\theta(1-\tan^2\theta)}{1-3\tan^2\theta}=\frac{4\tan\theta}{1-\tan^2\theta}\\ \Rightarrow 4\tan\theta\left[\frac{1-\tan^2\theta}{1-3\tan^2\theta}-\frac{1}{1-\tan^2\theta}\right]=0 \\ \Rightarrow 4\tan\theta[(1-\tan^2\theta)^2-1+3\tan^2\theta]=0 \\ \Rightarrow 4\tan\theta[1-2\tan^2\theta+\tan^4\theta\\-1+3\tan^2\theta]=0 \\ \Rightarrow 4\tan\theta[\tan^2\theta+\tan^4\theta]=0 \\ \Rightarrow 4\tan\theta. \tan^2\theta(1+\tan^2\theta)=0 \\ \Rightarrow \tan^3\theta=0,\,\,[1+\tan^2\theta\neq 0] \\ \Rightarrow \tan\theta=0 \\ \therefore \theta=n\pi,\,\,n\,\text{being any integer.}$

$\,(vii)\,\,\sin2\theta\tan\theta+1=\sin2\theta+\tan\theta\\ \Rightarrow \sin2\theta\tan\theta+1-\sin2\theta-\tan\theta=0\\ \Rightarrow \sin2\theta(\tan\theta-1)-1(\tan\theta-1)=0 \\ \Rightarrow (\tan\theta-1)(\sin2\theta-1)=0 \\ \therefore \tan\theta-1=0\rightarrow(1) \\ \text{and}\,\,\sin2\theta-1=0\rightarrow(2)$

From $\,(1)\,$ we get, $\,\,\tan\theta=1 \Rightarrow \theta=\frac{n\pi}{4}$

From $\,(2)\,\,$ we get, $\,\,\sin2\theta=1 \Rightarrow \theta=(4n+1)\frac{\pi}{4}$

Hence, the required general solution of the given euation is given by : 

$\,\,\theta=\frac{n\pi}{4},\,\,(4n+1)\frac{\pi}{4}\,\,,\,\,n\,\text{being any integer.}$

$\,(viii)\,\,\cos^3\theta\sin3\theta+\sin^3\theta\cos3\theta=\frac 34 \\ \Rightarrow 4\cos^3\theta\sin3\theta+4\sin^3\theta\cos3\theta=3 \\ \Rightarrow (\cos3\theta+3\cos\theta)\sin3\theta\\ +(3\sin\theta-\sin3\theta)\cos3\theta=3 \\ \Rightarrow \cos3\theta\sin3\theta+3\cos\theta\sin3\theta\\ +3\sin\theta\cos3\theta-\sin3\theta\cos3\theta=3 \\ \Rightarrow 3(\sin3\theta\cos\theta+\cos3\theta\sin\theta)=3 \\ \Rightarrow \sin(3\theta+\theta)=1\\ \Rightarrow \sin4\theta=1\\ \Rightarrow 4\theta=(4n+1)\frac{\pi}{2} \\ \Rightarrow \theta=(4n+1)\frac{\pi}{8},\,\,n\,\text{being any integer}$

$\,\,(ix)\,\,5 \cos\theta+2\sin\theta=2,\,\,$ given $\,\,\tan68^{\circ}12'=\frac 52.$

Sol. $\,\,5 \cos\theta+2\sin\theta=2 \\ \Rightarrow \frac{5}{\sqrt{5^2+2^2}}\cos\theta+\frac{2}{\sqrt{5^2+2^2}}\sin\theta=\frac{2}{\sqrt{5^2+2^2}} \\ \Rightarrow \frac{5}{\sqrt{29}}\cos\theta+\frac{2}{\sqrt{29}}\sin\theta=\frac{2}{\sqrt{29}}\rightarrow(1)$

We have, $\tan \alpha=\frac 52\,\,$ where $\,\,\alpha=68^{\circ}12'$

Now, $\,\,\tan\alpha=\frac{\sin\alpha}{\cos\alpha} \\ \Rightarrow \frac 52=\frac{\sin\alpha}{\cos\alpha} \\ \Rightarrow \frac{\sin\alpha}{5}=\frac{\cos\alpha}{2}=\frac{\sqrt{\sin^2\alpha+\cos^2\alpha}}{\sqrt{5^2+2^2}}=\frac{1}{\sqrt{29}} \\ \therefore \sin\alpha=\frac{5}{\sqrt{29}},\,\,\cos\alpha=\frac{2}{\sqrt{29}}$

Hence, from $\,(1)\,$ we get, 

 $\,\,\sin\alpha\cos\theta+\cos\alpha\sin\theta=\cos\alpha \\ \Rightarrow \sin(\theta+\alpha)=\sin(\frac{\pi}{2}-\alpha) \\ \Rightarrow \theta+\alpha=n\pi+(-1)^n\left(\frac{\pi}{2}-\alpha\right) \\ \Rightarrow \theta=n\pi+(-1)^n(90^{\circ}-68^{\circ}12')-68^{\circ}12' \\ \Rightarrow \theta=n\pi+(-1)^n21^{\circ}48'-68^{\circ}12'\,\,\text{(ans.)}$

$\,\,(x)\,\,4 \cos x+5 \sin x=5,\,\,$ given $\,\,\tan 51^{\circ}21'=\frac 54$

Sol. $\,\,4\cos x+5\sin x=5 \\ \Rightarrow \frac{4}{\sqrt{4^2+5^2}}\cos x+\frac{5}{\sqrt{4^2+5^2}}\sin x=\frac{5}{\sqrt{4^2+5^2}} \\ \Rightarrow \frac{4}{\sqrt{41}}\cos x+\frac{5}{\sqrt{41}}\sin x=\frac{5}{\sqrt{41}}\rightarrow(1)$

$\,\,\tan 51^{\circ}21'=\frac 54\,\,\text{(given)} \\ \Rightarrow \tan \alpha=\frac 54\,\,\,[\text{where}\,\,\,\alpha=51^{\circ}21'] \\ \Rightarrow \frac{\sin\alpha}{\cos\alpha}=\frac 54 \\ \Rightarrow \frac{\sin\alpha}{5}=\frac{\cos\alpha}{4}=\frac{\sqrt{\sin^2\alpha+\cos^2\alpha}}{\sqrt{5^2+4^2}}=\frac{1}{\sqrt{41}} \\ \therefore \sin\alpha=\frac{5}{\sqrt{41}},\,\,\cos\alpha=\frac{4}{\sqrt{41}}$

Hence, from $\,(1)\,$ we get, 

$\,\,\cos\alpha\cos x+\sin\alpha\sin x=\sin\alpha \\ \Rightarrow \cos(x-\alpha)=\cos(\pi/2-\alpha) \\ \Rightarrow x-\alpha=2n\pi \pm (\pi/2-\alpha) \\ \therefore x=2n\pi+\pi/2-\alpha+\alpha=2n\pi+\pi/2 \rightarrow(2) \\ \text{and}\,\,x=2n\pi-\pi/2+\alpha+\alpha\\~~~~~~~~~=2n\pi-\pi/2+2\alpha \\~~~~~~~~~=2n\pi+(2 \times 51^{\circ}21'-90^{\circ})\\~~~~~~~~~=2n\pi+12^{\circ}42'\rightarrow(3)$

From $\,\,(2),\,(3)\,\,$ we get the required general solution of the given equation.