Differentiation (Part-38) | S N De

 

$\,1(xiii)\,\,\sin5x\cos3x=\sin9x\cos7x \\ \Rightarrow 2\sin5x\cos3x=2\sin9x\cos7x \\ \Rightarrow\sin(5x+3x)+\sin(5x-3x)\\=\sin(9x+7x)+\sin(9x-7x) \\ \Rightarrow\sin8x+\sin2x=\sin16x+\sin2x \\ \Rightarrow\sin16x-\sin8x=0 \\ \Rightarrow 2\cos\left(\frac{16x+8x}{2}\right)\sin\left(\frac{16x-8x}{2}\right)=0 \\ \Rightarrow 2\cos12x\sin4x=0 \\ \therefore \cos12x=0 \rightarrow(1),\\ \text{or,}\,\,\sin4x=0 \rightarrow(2).$

From $\,(1)\,\,$ we get,  $\,\,12x=(2n+1)\pi/2,\,\,n\in\mathbb Z \\ \Rightarrow x=(2n+1)\frac{\pi}{24}.$

Again, by $\,(2)\,\,$ we get, $\,\,4x=n\pi \\ \Rightarrow x=\frac{n\pi}{4},\,\,n\in\mathbb Z.$

Hence, $\,\, x=(2n+1)\frac{\pi}{24},\,\,\frac{n\pi}{4},\,\,n\in\mathbb Z.$

$\,\,1(xiv)\,\,\cos9x\cos7x=\cos5x\cos3x,\, \\ -\frac{\pi}{4}\leq x\leq \frac{\pi}{4}. \\ \Rightarrow 2\cos9x\cos7x=2\cos5x\cos3x \\ \Rightarrow \cos(9x+7x)+\cos(9x-7x)\\=\cos(5x+3x)+\cos(5x-3x)\\ \Rightarrow \cos16x+\cos2x=\cos8x+\cos2x\\ \Rightarrow \cos16x-\cos8x=0 \\ \Rightarrow 2\sin12x\sin4x=0 \,\,[*] \\ \therefore \sin12x=0\rightarrow(1),\\ \text{or,}\,\,\,\sin4x=0\rightarrow(2).$

From $\,(1)\,\,$ we get, $\,\,12x=n\pi \\ \Rightarrow x=\frac{n\pi}{12},\,\,n \in\mathbb Z\rightarrow(3)$

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

By $\,(3)\,$ we get, for$\,\,n=0,\,\,x=0 $

By $\,(3)\,$ we get, for$\,\,n=\pm 1,\,\,x=\pm \pi/12 $

By $\,(3)\,$ we get, for$\,\,n=\pm2,\,\,x=\pm \pi/6 $

By $\,(3)\,$ we get, for$\,\,n=\pm3,\,\,x=\pm \pi/4 $

By $\,(3)\,$ we get, for$\,\,n=\pm4,\,\,x=\pm \pi/3 $

By $\,(4)\,$ we get, for$\,\,n=0,\,\,x=0 $

By $\,(4)\,$ we get, for$\,\,n=\pm 1,\,\,x=\pm \pi/4 $

By $\,(4)\,$ we get, for$\,\,n=\pm 2,\,\,x=\pm \pi/2 $

Now, since $\,\, -\pi/4 \leq x \leq \pi/4 \\ \therefore x=0, \pm \pi/12, \pm \pi/6, \pm \pi/4.$

Note[*] : $\cos(a)-\cos(b)=-2\sin\left(\frac{a+b}{2}\right)\sin\left(\frac{a-b}{2}\right)$

$\,1(xv)\,\,2\sin^2\theta+3\cos\theta=0 \\ \Rightarrow 2-2\cos^2\theta+3\cos\theta=0 \\ \Rightarrow 2\cos^2\theta-3\cos\theta-2=0 \\ \Rightarrow 2\cos^2\theta-4\cos\theta+\cos\theta-2=0 \\ \Rightarrow 2\cos\theta(\cos\theta-2)+1(\cos\theta-2)=0 \\ \Rightarrow (\cos\theta-2)(2\cos\theta+1)=0 \\ \therefore \cos\theta=2,-\frac 12  \\ \text{but since},\,\,-1\leq \cos\theta\leq 1,\\ \therefore \cos\theta \neq 2 \\ \text{and so}, \cos\theta=-\frac 12=\cos\frac{2\pi}{3} \\ \Rightarrow \theta=2n\pi \pm 2\pi/3,\,\,n\in\mathbb Z.$

$\,1(xvi)\,\,\cos^2\theta-\sin\theta=\frac 14\,\,\,\left(0^{\circ}<\theta<360^{\circ}\right) \\ \Rightarrow 1-\sin^2\theta-\sin\theta=\frac 14 \\ \Rightarrow 4\sin^2\theta+4\sin\theta-3=0 \\ \Rightarrow 4\sin^2\theta+6\sin\theta-2\sin\theta-3=0 \\ \Rightarrow 2\sin\theta(2\sin\theta+3)-1(2\sin\theta+3)=0 \\ \Rightarrow (2\sin\theta+3)(2\sin\theta-1)=0 \\ \Rightarrow 2\sin\theta-1=0 \\ [\,\,\text{since}\,\,\,2\sin\theta+3 \neq 0,\,\,\text{as}\,\, -1 \leq\sin\theta\leq 1] \\ \Rightarrow \sin\theta=\frac 12\\ \Rightarrow \theta=30^{\circ},150^{\circ} \,\,\left(0^{\circ}<\theta<360^{\circ}\right) $

$\,1(xvii)\,\,2\cos^2\theta -\sin\theta +1=0 \\~~~~~~~~~~~~~~~~~~~~~~~~ \left(0^{\circ}<\theta <1000^{\circ}\right) \\ \Rightarrow 2-2\sin^2\theta -\sin\theta +1=0 \\ \Rightarrow 2\sin^2\theta +\sin\theta -3=0 \\ \Rightarrow 2\sin^2\theta +3\sin\theta -2\sin\theta -3=0 \\ \Rightarrow \sin\theta (2\sin\theta +3)-1(2\sin\theta +3)=0 \\ \Rightarrow (2\sin\theta +3)(\sin\theta -1)=0 \\ \therefore 2\sin\theta +3=0 \Rightarrow \sin\theta =-3/2,\\\text{which is not possible.} \\ \text{so,}\,\,\sin\theta -1=0 \\ \Rightarrow \sin\theta =1 \\ \therefore \theta =90^{\circ},450^{\circ},810^{\circ}\,\,\left(0^{\circ}<\theta <1000^{\circ}\right)$

$\,1(xviii)\,\,\cos2x-5\cos x+3=0 \\~~~~~~~~~~~~~~~~~~~~~~~~\left(0<x<2\pi\right) \\ \Rightarrow 2\cos^2x-1-5\cos x+3=0  \\ \Rightarrow 2\cos^2x-5\cos x+2=0  \\ \Rightarrow 2\cos^2x-4\cos x-\cos x+2=0  \\ \Rightarrow 2\cos x(\cos x-2)-1(\cos x-2)=0  \\ \Rightarrow(\cos x-2)(2\cos x-1)=0 \\ \therefore \cos x=2,\frac 12 \\ \text{hence,}\,\,\cos x=\frac 12\,\,[\,\text{as}\,\cos x \neq 2]  \\ \Rightarrow x=\pi/3, \,\,5\pi/3\,\,\left(0<x<2\pi\right)$

$\,1(xix)\,\,3(\sec^2\theta +\tan^2\theta )=5 \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\left(0^{\circ}<\theta <360^{\circ} \right) \\ \Rightarrow 3(1+\tan^2\theta +\tan^2\theta )=5 \\ \Rightarrow 3(1+2\tan^2\theta )=5 \\ \Rightarrow 6\tan^2\theta =5-3 \\ \Rightarrow \tan^2\theta =\frac 26=\frac 13 \\ \Rightarrow \tan\theta =\pm\frac{1}{\sqrt3} \\ \Rightarrow \theta =30^{\circ},150^{\circ},210^{\circ},330^{\circ} \,\,\left(0^{\circ}<\theta <360^{\circ} \right)$

$\,1(xx)\,\,2(\sec^2\theta+\sin^2\theta)=5 \\ \Rightarrow \sec^2\theta+1-\cos^2\theta=\frac 52 \\ \Rightarrow 1/x^2+1-x^2=\frac 52\,\,\text{where,}\,\,x=\cos\theta \\ \Rightarrow 2(1+x^2-x^4)=5x^2 \\ \Rightarrow 2x^4+3x^2-2=0 \\ \Rightarrow 2x^4+4x^2-x^2-2=0 \\ \Rightarrow 2x^2(x^2+2)-1(x^2+2)=0 \\ \Rightarrow (x^2+2)(2x^2-1)=0 \\ \Rightarrow 2x^2-1=0\,\,[x^2+2\neq 0,\,\,\text{as}\,\,-1\leq x\leq 1]\\ \Rightarrow x=\pm \frac{1}{\sqrt2}\rightarrow(1)$

By $\,(1)\,$ we get, $\,\,\cos\theta=1/\sqrt2=\cos\frac{\pi}{4} \\ \Rightarrow \theta=2n\pi \pm \pi/4,\,\,n\in\mathbb Z\rightarrow(2)$

Again, from $\,(1)\,$ we get, $\,\,\cos\theta=-1/\sqrt2=\cos\frac{3\pi}{4} \\ \Rightarrow \theta=2n\pi \pm \frac{3\pi}{4},\,\,n\in\mathbb Z\rightarrow(3)$

Hence, from $\,\,(2)\, , (3)\, \,\,$ we get the required general solution.