Differentiation (Part-38) | S N De

 

$\,\,1(xxi)\,\cos\theta -\sin\theta =\frac{1}{\sqrt2} \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~(-\pi<\theta<\pi) \\ \Rightarrow \frac{1}{\sqrt2}\cos\theta -\frac{1}{\sqrt2}\sin\theta =\frac 12 \\ \Rightarrow \sin\left(\frac{\pi}{4}-\theta \right)=\sin\frac{\pi}{6} \\ \Rightarrow \pi/4-\theta =n\pi+(-1)^n\pi/6,\,\,n\in\mathbb Z \\ \Rightarrow \theta =\pi/4-n\pi-(-1)^n\pi/6\rightarrow(1)$

From $\,(1)\,\,$ we get,  for $\,n=0\,,\,\,\theta =\pi/12$

Again from $\,(1)\,\,$ we get,  for $\,n=1\,,\,\,\theta =-7\pi/12,\,\,\,17\pi/12$

So, $\,\,\theta=\pi/12,\,\,-7\pi/12\,\,\,\text{since,}\,\, -\pi<\theta<\pi.$

$\,1(xxii)\,\,\sin\theta-\sqrt3\cos\theta=1 \\ \Rightarrow \frac 12\sin\theta-\frac{\sqrt3}{2}\cos\theta=\frac 12 \\ \Rightarrow \sin(\pi/6)\sin\theta-\cos(\pi/6)\cos\theta=\frac12\\ \Rightarrow -\cos(\theta+\pi/6)=\frac12 \\ \Rightarrow \cos(\theta+\pi/6)=\cos(2\pi/3) \\ \Rightarrow \theta+\pi/6=2n\pi\pm 2\pi/3 \\ \therefore \theta=2n\pi+\pi/2,\,\,2n\pi-5\pi/6,\,\,n\in\mathbb Z.$

$\,1(xxiii)\,\, \sin x+\cos x=1/\sqrt2 \\ \Rightarrow \frac{1}{\sqrt2} \cos x+\frac{1}{\sqrt2}\sin x=\frac 12\\ \Rightarrow  \cos (x-\pi/4)=\cos(\pi/3) \\ \Rightarrow x-\pi/4=2n\pi \pm \pi/3 \\ \therefore x=2n\pi+\frac{7\pi}{12},\,\,2n\pi-\frac{\pi}{12},\,\,n\in\mathbb Z$

$\,1(xxiv)\,\,\sqrt3\sin x+\cos x=\sqrt2\\ \Rightarrow  \frac{\sqrt3}{2}\sin x+\frac 12 \cos x=\frac{\sqrt2}{2}\\ \Rightarrow \frac12 \cos x+(\sqrt3/2) \sin x=\frac{1}{\sqrt2}\\ \Rightarrow \cos(x-\pi/3)=\cos(\pi/4)\\ \Rightarrow  x-\pi/3=2n\pi \pm \pi/4,\,\,n\in\mathbb Z\\ \Rightarrow x=2n\pi+7\pi/12,\,\,\,2n\pi+\pi/12\,\,\text{(ans.)}$

$\,1(xxv)\,\,\sin2\theta-\cos2\theta=1 \\ \Rightarrow \frac{1}{\sqrt2}\sin2\theta-\frac{1}{\sqrt2}\cos2\theta=\frac{1}{\sqrt2} \\ \Rightarrow -\left(\cos\frac{\pi}{4}\cos2\theta-\sin\frac{\pi}{4}\sin2\theta\right)=\frac{1}{\sqrt2} \\ \Rightarrow \cos\left(2\theta+\frac{\pi}{4}\right)=-\frac{1}{\sqrt2}=\cos\frac{3\pi}{4} \\ \therefore 2\theta+\frac{\pi}{4}=2n\pi \pm\frac{3\pi}{4} \\ \Rightarrow 2\theta=2n\pi+\pi/2,\,\,2n\pi-\pi \\ \Rightarrow \theta=n\pi+\pi/4,\,\,n\pi-\pi/2,\quad\,n\in\mathbb Z.$

$\,1(xxvi)\,\,\sin\theta+2\cos\theta=1 \\ \Rightarrow \frac{1}{\sqrt{1^2+2^2}}\sin\theta+\frac{2}{\sqrt{1^2+2^2}}\cos\theta=\frac{1}{\sqrt{1^2+2^2}} \\ \Rightarrow \frac{1}{\sqrt5}\sin\theta+\frac{2}{\sqrt5}\cos\theta=\frac{1}{\sqrt5}\\ \Rightarrow \sin\alpha\sin\theta+\cos\alpha\cos\theta=\sin\alpha \\ \text{let,}\,\,\frac{1}{\sqrt5}=\sin\alpha,\,\,\left(0<\alpha<\frac{\pi}{2}\right) \\ \Rightarrow \cos(\theta-\alpha)=\cos\left(\frac{\pi}{2}-\alpha\right) \\ \Rightarrow \theta-\alpha=2n\pi \pm \left(\pi/2-\alpha\right) \\ \therefore \theta=2n\pi+\pi/2,\,\,2n\pi-\pi/2+2\alpha \\ \text{where}\,\,n\in\mathbb Z,\,\,\frac{1}{\sqrt5}=\sin\alpha,\,\,\left(0<\alpha<\frac{\pi}{2}\right)$

$\,1(xxvii)\,\,\,\tan^2x-(1+\sqrt3)\tan x+\sqrt3=0 \\ \Rightarrow (\tan x-\sqrt3)(\tan x-1)=0 \\ \Rightarrow  \tan x-\sqrt3=0 \rightarrow(1), \\ \text{or},\,\,\, \tan x-1=0\rightarrow(2).$

From $\,(1)\,\,$ we get, $\,\,\,\tan x=\sqrt3 \Rightarrow x=n\pi+\pi/3$

and from $\,(2)\,\,$ we get, $\,\,\,\tan x=1 \Rightarrow x=n\pi+\pi/4$

Hence, $\,\,x=n\pi+\pi/3,\,\,n\pi+\pi/4,\,\,\,n\in\mathbb Z.$

$\,1(xxviii)\,\,\, \tan x+\tan 2x+\tan x\tan 2x=1 \\ \Rightarrow  \tan x+\tan2x=1-\tan x\tan2x \\ \Rightarrow \frac{\tan x+\tan2x}{1-\tan x\tan2x}=1\\ \Rightarrow \tan(x+2x)=1 \\ \Rightarrow \tan3x=1 \\ \Rightarrow 3x=n\pi+\pi/4 \\ \Rightarrow  x=n\pi/3+\pi/12 \\ \Rightarrow x=(4n+1)\pi/12,\,\,\,\,n\in\mathbb Z.$

$\,1(xxix)\,\,\tan \theta +\tan2\theta \\ +\sqrt3\tan\theta \tan2\theta =\sqrt3 \\ \Rightarrow \tan\theta +\tan2\theta =\sqrt3(1-\tan\theta \tan2\theta ) \\ \Rightarrow \frac{\tan\theta +\tan2\theta }{1-\tan\theta \tan2\theta }=\sqrt3 \\ \Rightarrow \tan(\theta +2\theta )=\tan (\pi/3) \\ \Rightarrow \tan 3\theta =\tan(\pi/3) \\ \Rightarrow 3\theta =n\pi+\pi/3 \\ \Rightarrow \theta =\frac{n\pi}{3}+\frac{\pi}{9} \\ \Rightarrow \theta =(3n+1)\pi/9,\,\,\,n \in\mathbb Z.$

$\,1(xxx)\,\,\tan\left(\pi/4+\theta\right)+\tan\left(\pi/4-\theta\right)=4 \\ \Rightarrow \frac{1+\tan\theta}{1-\tan\theta}+\frac{1-\tan\theta}{1+\tan\theta}=4 \\ \Rightarrow \frac{(1+\tan\theta)^2+(1-\tan\theta)^2}{(1-\tan\theta)(1+\tan\theta)}=4 \\ \Rightarrow \frac{2(1+\tan^2\theta)}{1-\tan^2\theta}=4 \\ \Rightarrow \frac{1-\tan^2\theta}{1+\tan^2\theta}=\frac 12 \\ \Rightarrow \cos2\theta=\frac 12 =\cos(\pi/3)\\ \Rightarrow 2\theta=2n\pi \pm \pi/3 \\ \Rightarrow \theta=n\pi \pm \pi/6,\,\,\, n\in\mathbb Z.$

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