Differentiation (Part-38) | S N De

 

$\,1(xxxi)\,\,\frac{\sin\alpha}{\sin2x}+\frac{\cos\alpha}{\cos2x}=2 \\ \Rightarrow \frac{\sin\alpha\cos2x+\cos\alpha\sin2x}{\sin2x\cos2x}=2 \\ \Rightarrow \sin(\alpha+2x)=2\sin2x\cos2x \\ \Rightarrow \sin(\alpha+2x)=\sin4x \\ \Rightarrow \sin4x-\sin(\alpha+2x)=0 \\ \Rightarrow 2\cos(3x+\alpha/2)\sin(x-\alpha/2)=0 \\ \therefore \cos(3x+\alpha/2)=0\rightarrow(1),\\ \text{or,}\,\,\, \sin(x-\alpha/2)=0 \rightarrow(2)$

From $\,(1),\,\,$ we get, $\,\,3x+\alpha/2=(2n+1)\pi/2 \\ \Rightarrow x=(2n+1)\frac{\pi}{6}-\frac{\alpha}{6},\,\,\, n\in\mathbb Z\rightarrow(3)$

Again, by $\,(2),\,\,$ we get, $\,\,x-\alpha/2=n\pi \\ \Rightarrow x=n\pi+\alpha/2,\,\,n\in\mathbb Z\rightarrow(4)$

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

$\,1(xxxii)\,\,\sqrt3\tan\theta\tan(\theta+\pi/3) \\ \times \tan(\theta+2\pi/3)=1 \\ \Rightarrow \tan\theta . \frac{\tan\theta+\sqrt3}{1-\sqrt3\tan\theta}.\frac{\tan\theta-\sqrt3}{1+\sqrt3\tan\theta}=\frac{1}{\sqrt3} \\ \Rightarrow \tan\theta. \frac{\tan^2\theta-3}{1-3\tan^2\theta}=\frac{1}{\sqrt3} \\ \Rightarrow \frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}=-\frac{1}{\sqrt3} \\ \Rightarrow \tan 3\theta=\tan\left(-\frac{\pi}{6}\right) \\ \Rightarrow 3\theta=n\pi-\pi/6 \\ \Rightarrow \theta=n\pi/3-\pi/18,\,\,\, n\in\mathbb Z.$

$\,1(xxxiii)\,\,\,(2+\sqrt3)\cos\theta=1-\sin\theta \\ \Rightarrow \sin\theta+(2+\sqrt3)\cos\theta=1 \\ \Rightarrow \frac{1}{2\sqrt{2+\sqrt3}} \sin\theta+\frac{2+\sqrt3}{2\sqrt{2+\sqrt3}}\cos\theta=\frac{1}{2\sqrt{2+\sqrt3}} \\ \Rightarrow \cos(\theta-\alpha)=\sin\alpha \\ \text{let,}\,\,\frac{1}{2\sqrt{2+\sqrt3}}=\sin\alpha \\ \Rightarrow \cos(\theta-\alpha)=\cos(\pi/2-\alpha) \\ \Rightarrow \theta-\alpha=2n\pi \pm (\pi/2-\alpha) \\ \therefore \theta=2n\pi+\pi/2, \,\,2n\pi-\pi/2+2\alpha \\ \text{where}\,\,\,\frac{1}{2\sqrt{2+\sqrt3}}=\sin\alpha,\,\, n \in\mathbb Z.$

$\,1(xxxiv)\,\,\,\cos2\theta=(\sqrt2+1)\left(\cos\theta-\frac{1}{\sqrt2}\right) \\ \Rightarrow 2\cos^2\theta-1=(\sqrt2+1)\cos\theta-1-\frac{1}{\sqrt2} \\ \Rightarrow 2\cos^2\theta-(\sqrt2+1)\cos\theta+\frac{1}{\sqrt2}=0 \\ \Rightarrow 2\sqrt2\cos^2\theta-(2+\sqrt2)\cos\theta+1=0 \\ \Rightarrow 2\cos\theta(\sqrt2\cos\theta-1)-1(\sqrt2\cos\theta-1)=0 \\ \Rightarrow (\sqrt2\cos\theta-1)(2\cos\theta-1)=0 \\ \therefore \sqrt2\cos\theta-1=0 \rightarrow(1), \\ \text{or},\,\,\,\, 2\cos\theta-1=0\rightarrow(2)$

From $\,(1)\,\, $, we get, $\,\,\cos\theta=1/\sqrt2 =\cos(\pm \pi/4)\\ \Rightarrow \theta=2n\pi \pm \pi/4,\,\,n\in\mathbb Z \rightarrow(3)$

By $\,(2)\,\,$ we get, $\,\,\cos\theta=\frac 12=\cos(\pm \pi/3) \\ \Rightarrow \theta=2n\pi \pm \pi/3,\,\, n\in\mathbb Z \rightarrow(4)$

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

 $\,1(xxxv)\,\,\,\cos3\theta+\cos2\theta\\=\sin(3\theta/2)+\sin(\theta/2) \,\,\,\, \text{where}\,\,0 \leq \theta \leq 2\pi \\ \Rightarrow 2\cos\left(\frac{3\theta+2\theta}{2}\right) \cos\left(\frac{3\theta-2\theta}{2}\right)=2\sin\theta\cos\frac{\theta}{2} \\ \Rightarrow 2\cos(5\theta/2)\cos(\theta/2)=2\sin\theta\cos(\theta/2) \\ \Rightarrow 2\cos(\theta/2)\left[\cos\frac{5\theta}{2}-\sin\theta\right]=0 \\ \Rightarrow \cos(\theta/2)=0 \rightarrow(1),\\ \text{or},\,\, \cos(5\theta/2)-\sin\theta=0\rightarrow(2).$

From $\,(1)\,\,$, we get, $\,\,\theta/2=(2n+1)\frac{\pi}{2} \\ \Rightarrow \theta=(2n+1)\pi\rightarrow(3)$

Again, by $\,(2),\,\,$ we get, $\,\,\cos(5\theta/2)=\cos(\pi/2-\theta) \\ \Rightarrow 5\theta/2=2n\pi \pm \left(\frac{\pi}{2}-\theta\right) \\ \therefore \theta=(4n+1)\pi/7,\rightarrow(4) \\ \text{and}\,\, \theta=(4n-1)\pi/3 \rightarrow(5)$

By $\,(3)\,$ we get, $\,\,\theta=\pi\,\, \text{for}\,(n=0) \\ \text{and}\,\, \theta=3\pi,\,\,-\pi\,\,\text{for}\,\,(n=\pm 1)$

Again, by $\,(4)\,\,$, we get, $\,\,\theta=\pi/7\,\,(n=0), \\ \theta=5\pi/7,\, -3\pi/7\,\,(n=\pm1) \\ \theta= 9\pi/7,\,\, -\pi\,\,(n=\pm 2) \\ \theta=13\pi/7,\,\,-11\pi/7\,\,(n=\pm3) \\ \theta=17\pi/7,\,\,-15\pi/7\,\,(n=\pm4)$

Finally , by $\,(5)\,\,$, we get, $\,\,\theta=-\pi/3,\,\,(n=0) \\ \theta=\pi,\,-5\pi/3\,\,(n=\pm1)$

$\,\therefore \,\, \theta=\pi,\,\pi/7,\,5\pi/7,\,9\pi/7,\,13\pi/7 \\ [0 \leq \theta \leq 2\pi]$

$\,1(xxxvi)\,\cos3x+\sin\left(2x-7\pi/6\right)=-2 \\ \therefore \cos3x=-1\rightarrow(1), \\ \text{and}\,\,\sin(2x-7\pi/6)=-1\rightarrow(2)\\ \text{since},\,\,-1\leq \sin\theta,\cos\theta \leq 1 $

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

Again, by $\,(2)\,$ we get, $\,\,2x-7\pi/6=(4n-1)\pi/2 \\ \Rightarrow x=n\pi+\frac{\pi}{3},\,n\in\mathbb Z$

Hence, both $\,(1)\,$ and $\,(2)\,$ will satisfy if  $\,\,x=2n\pi+\frac{\pi}{3},\,n\in\mathbb Z.$

$\,1(xxxvii)\,\,\sin x+\cos2x=-2 \\ \Rightarrow \sin x=-1 \rightarrow(1),\\ \text{and}\,\,\cos2x=-1 \rightarrow(2)\\ \text{since},\,\,-1 \leq \sin \theta,\cos \theta \leq 1$

Now, from $\,(1)\,\,$, we get, $\,\,x=(4n-1)\pi/2 ,\,n\in\mathbb Z \rightarrow(3)$

Again, by $\,(2)\,$, we get, $\,\,2x=(2n+1)\pi \\ \Rightarrow x=(2n+1)\pi/2,\,\, n\in\mathbb Z\rightarrow(4).$

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

To continue with GENERAL SOLUTIONS OF TRIGONOMETRIC EQUATIONS (PART-13) , click here .