Differentiation (Part-38) | S N De

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

Solution

$4\sin\theta\cos\theta=1+2\cos\theta-2\sin\theta$

$\Rightarrow 4\sin\theta\cos\theta+2\sin\theta-2\cos\theta-1=0$

$\Rightarrow 2\sin\theta(2\cos\theta+1)-1(2\cos\theta+1)=0$

$\Rightarrow (2\cos\theta+1)(2\sin\theta-1)=0$

$\therefore 2\cos\theta+1=0\rightarrow(1), \quad 2\sin\theta-1=0\rightarrow(2)$

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

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

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

Solution

$1+2\sin\theta\cos\theta-2\sin\theta-\cos\theta=0$

$\Rightarrow 1-\cos\theta-2\sin\theta(1-\cos\theta)=0$

$\Rightarrow (1-\cos\theta)-2\sin\theta(1-\cos\theta)=0$

$\Rightarrow (1-\cos\theta)(1-2\sin\theta)=0$

$\therefore 1-\cos\theta=0 \rightarrow(1) \quad \text{and}\,\,1-2\sin\theta=0\rightarrow(2)$

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

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

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

Solution

$1-2\sin\theta-2\cos\theta+\frac{\cos\theta}{\sin\theta}=0$

$\Rightarrow \sin\theta-2\sin^2\theta-2\sin\theta\cos\theta+\cos\theta=0$

$\Rightarrow \sin\theta(1-2\sin\theta)+\cos\theta(1-2\sin\theta)=0$

$\Rightarrow (1-2\sin\theta)(\sin\theta+\cos\theta)=0$

$\therefore 1-2\sin\theta=0 \rightarrow(1), \quad \sin\theta+\cos\theta=0 \rightarrow(2)$

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

$(iv).\,\tan x+\tan 2x+\tan3x=0$

Solution

$\tan x+\tan2x+\tan(x+2x)=0$

$\Rightarrow \tan x+\tan2x+\frac{\tan x+\tan 2x}{1-\tan x\tan2x}=0$

$\Rightarrow (\tan x+\tan2x)\left(1+\frac{1}{1-\tan x\tan2x}\right)=0$

$\Rightarrow (\tan x+\tan2x)(2-\tan x\tan2x)=0$

$\Rightarrow(\tan x+\tan2x)\left(2-\tan x.\frac{2\tan x}{1-\tan^2x}\right)=0$

$\Rightarrow (\tan x+\tan2x) \times \left[2(1-\tan^2x)-2\tan^2x\right]=0$

$\Rightarrow(\tan x+\tan2x)(2-4\tan^2x)=0$

$\therefore (\tan x+\tan2x)=0\rightarrow(1), \quad \text{and}\,\,0\,(2-4\tan^2x)=0\rightarrow(2)$

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

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

$(v)\,\,\,\tan x+\tan2x+\tan3x=\tan x\tan2x\tan3x$

Solution

$\tan x+\tan2x=-\tan3x+\tan x\tan2x\tan3x$

$\Rightarrow \tan x+\tan2x=-\tan3x(1-\tan x\tan2x)$

$\Rightarrow \frac{\tan x+\tan 2x}{1-\tan x\tan2x}=-\tan3x \Rightarrow \tan(x+2x)=-\tan3x$

$\Rightarrow 2\tan3x=0 \Rightarrow \tan3x=0 \Rightarrow 3x=n\pi,(\,\,n\,\text{being any integer})$

$\therefore x=\frac{n\pi}{3}$

$(vi)\,\,\tan3\theta+\tan\theta=2\tan2\theta$

Solution

$\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}+\tan\theta=2.\frac{2\tan\theta}{1-\tan^2\theta}$

$\Rightarrow \frac{3\tan\theta-\tan^3\theta+\tan\theta-3\tan^3\theta}{1-3\tan^2\theta}=\frac{4\tan\theta}{1-\tan^2\theta}$

$\Rightarrow \frac{4\tan\theta(1-\tan^2\theta)}{1-3\tan^2\theta}=\frac{4\tan\theta}{1-\tan^2\theta}$

$\Rightarrow 4\tan\theta\left[\frac{1-\tan^2\theta}{1-3\tan^2\theta}-\frac{1}{1-\tan^2\theta}\right]=0$

$\Rightarrow 4\tan\theta[(1-\tan^2\theta)^2-1+3\tan^2\theta]=0$

$\Rightarrow 4\tan\theta[1-2\tan^2\theta+\tan^4\theta-1+3\tan^2\theta]=0$

$\Rightarrow 4\tan\theta[\tan^2\theta+\tan^4\theta]=0 \Rightarrow 4\tan\theta. \tan^2\theta(1+\tan^2\theta)=0$

$\Rightarrow \tan^3\theta=0,\,\,[1+\tan^2\theta\neq 0] \Rightarrow \tan\theta=0$

$\therefore \theta=n\pi,\,\,n\,\text{being any integer.}$

$(vii)\,\,\sin2\theta\tan\theta+1=\sin2\theta+\tan\theta$

Solution

$\sin2\theta\tan\theta+1-\sin2\theta-\tan\theta=0$

$\Rightarrow \sin2\theta(\tan\theta-1)-1(\tan\theta-1)=0 \Rightarrow (\tan\theta-1)(\sin2\theta-1)=0$

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

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

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

Hence, the required general solution is: $\theta=\frac{n\pi}{4},\,\,(4n+1)\frac{\pi}{4}\,\,,\,\,n\,\text{being any integer.}$

$(viii)\,\,\cos^3\theta\sin3\theta+\sin^3\theta\cos3\theta=\frac 34$

Solution

$4\cos^3\theta\sin3\theta+4\sin^3\theta\cos3\theta=3$

$\Rightarrow (\cos3\theta+3\cos\theta)\sin3\theta+(3\sin\theta-\sin3\theta)\cos3\theta=3$

$\Rightarrow \cos3\theta\sin3\theta+3\cos\theta\sin3\theta+3\sin\theta\cos3\theta-\sin3\theta\cos3\theta=3$

$\Rightarrow 3(\sin3\theta\cos\theta+\cos3\theta\sin\theta)=3 \Rightarrow \sin(3\theta+\theta)=1$

$\Rightarrow \sin4\theta=1 \Rightarrow 4\theta=(4n+1)\frac{\pi}{2}$

$\therefore \theta=(4n+1)\frac{\pi}{8},\,\,n\,\text{being any integer.}$

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

Solution

$\frac{5}{\sqrt{5^2+2^2}}\cos\theta+\frac{2}{\sqrt{5^2+2^2}}\sin\theta=\frac{2}{\sqrt{5^2+2^2}}$

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

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

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

$\therefore \sin\alpha=\frac{5}{\sqrt{29}},\,\,\cos\alpha=\frac{2}{\sqrt{29}}$

Hence, from $\,(1)\,$ we get, $\,\,\sin\alpha\cos\theta+\cos\alpha\sin\theta=\cos\alpha \Rightarrow \sin(\theta+\alpha)=\sin(\frac{\pi}{2}-\alpha)$

$\Rightarrow \theta+\alpha=n\pi+(-1)^n\left(\frac{\pi}{2}-\alpha\right)$

$\Rightarrow \theta=n\pi+(-1)^n(90^{\circ}-68^{\circ}12')-68^{\circ}12'$

$\therefore \theta=n\pi+(-1)^n21^{\circ}48'-68^{\circ}12'\,\,\text{(ans.)}$

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

Solution

$\frac{4}{\sqrt{4^2+5^2}}\cos x+\frac{5}{\sqrt{4^2+5^2}}\sin x=\frac{5}{\sqrt{4^2+5^2}}$

$\Rightarrow \frac{4}{\sqrt{41}}\cos x+\frac{5}{\sqrt{41}}\sin x=\frac{5}{\sqrt{41}}\rightarrow(1)$

Let $\tan \alpha=\frac 54\,\,\,[\text{where}\,\,\,\alpha=51^{\circ}21'] \Rightarrow \frac{\sin\alpha}{\cos\alpha}=\frac 54$

$\Rightarrow \frac{\sin\alpha}{5}=\frac{\cos\alpha}{4}=\frac{\sqrt{\sin^2\alpha+\cos^2\alpha}}{\sqrt{5^2+4^2}}=\frac{1}{\sqrt{41}} \Rightarrow \sin\alpha=\frac{5}{\sqrt{41}},\,\,\cos\alpha=\frac{4}{\sqrt{41}}$

Hence, from $\,(1)\,$ we get, $\,\,\cos\alpha\cos x+\sin\alpha\sin x=\sin\alpha \Rightarrow \cos(x-\alpha)=\cos(\pi/2-\alpha)$

$\Rightarrow x-\alpha=2n\pi \pm (\pi/2-\alpha) \Rightarrow x=2n\pi+\pi/2-\alpha+\alpha=2n\pi+\pi/2 \rightarrow(2)$

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.