Differentiation (Part-38) | S N De

 

$\,6.\,$ If $\,\,16\tan^8\phi=\cos2\theta+3\cos\theta\,\,$ and $\,\, 3\cos^2\phi=2,\,\,$ then find the general values of $\,\,\theta.$

Sol. $\,\,\cos^2\phi=2\quad \text{(given)} \\ \Rightarrow \cos^2\phi=\frac 23 \\ \therefore \tan^2\phi=\frac{\sin^2\phi}{\cos^2\phi}=\frac{1-2/3}{2/3}=\frac 12. \\ \therefore 16 \times  (1/2)^4=\cos2\theta+3\cos\theta \\ \Rightarrow 1=2\cos^2\theta-1+3\cos\theta \\ \Rightarrow 2\cos^2\theta+3\cos\theta-2=0 \\ \Rightarrow 2\cos^2\theta+4\cos\theta-\cos\theta-2=0 \\ \Rightarrow 2\cos\theta(\cos\theta+2)-1(\cos\theta+2)=0 \\ \Rightarrow (\cos\theta+2)(2\cos\theta-1)=0 \\ \Rightarrow 2\cos\theta-1=0\,\,[**] \\ \Rightarrow \cos\theta=\frac 12 \\ \therefore \theta=2n\pi \pm \frac{\pi}{3},\,\,n \,\,\text{being any integer.}$

Note [**] : $\,\cos\theta+2\neq 0\,\,$ as $\,\,-1 \leq \cos\theta\leq 1.$

$\,\,7.\,\,$ Solve : 

$\,\,7.\,\,$ Solve : 

$\,\,2(\sin x-\cos2x)-\sin2x(1+2\sin x)\\+2\cos x=0 \\ \Rightarrow 2 \left[\sin x-(1-2\sin^2x)\right]\\-2\sin x\cos x(1+2\sin x)+2\cos x=0 \\ \Rightarrow 2\sin x-2+4 \sin^2x\\-2\sin x\cos x-4 \sin^2x \cos x+2\cos x=0 \\ \Rightarrow 2\sin x (1-\cos x)+4\sin^2x (1-\cos x)\\-2(1-\cos x)=0 \\ \Rightarrow (1-\cos x)(2\sin x+4\sin^2x-2)=0 \\ \Rightarrow (1-\cos x) \times 2(\sin x+2\sin^2x-1)=0 \\ \Rightarrow (1-\cos x)(2\sin^2x+2\sin x\\-\sin x-1)=0 \\ \Rightarrow (1-\cos x)\left[2\sin x(\sin x+1)\\-1(\sin x+1)\right]=0 \\ \Rightarrow (1-\cos x)(\sin x+1)(2\sin x-1)\\=0 \rightarrow(1)$

From $\,(1),\,\,$ we get, $\,\,1-\cos x=0 \\ \Rightarrow \cos x=1 \\ \Rightarrow x=2n\pi\rightarrow(2)$

Again, by $\,(1),\,\,$ we have, $\,\,\sin x+1=0 \\ \Rightarrow \sin x=-1 \\ \Rightarrow x=(4n-1)\pi/2\rightarrow(3)$

and  $\,\,2\sin x-1=0 \\ \Rightarrow \sin x=\frac 12 \\ \therefore x=n\pi+(-1)^n \pi/6\rightarrow(4)$

Hence, from $\,\,(2)\, , (3),\, (4)\,\,$ we get the required general solution. 

$\,8.\,\,$ Prove that, the system of equations $\,\,x-y=2\pi/3\,\,$ and $\,\,\cos x+\cos y=\frac 32\,\,$ does not have any real solution.

Sol. We have , $\,\,x-y=2\pi/3\rightarrow(1)$ 

and $\,\,\cos x+\cos y=\frac 32\rightarrow (2) \\ \Rightarrow 2 \cos \frac{x+y}{2}\cos\frac{x-y}{2}=\frac 32 \\ \Rightarrow 2 \cos\frac{x+y}{2}\times  \cos \left(\frac 12 .\frac{2\pi}{3}\right)=\frac 32 \\ \Rightarrow 2 \cos\frac{x+y}{2} \times \cos\frac{\pi}{3}=\frac 32 \\ \Rightarrow 2\cos\frac{x+y}{2} \times \frac 12=\frac 32 \\ \therefore \cos\frac{x+y}{2}=\frac 32\,\,\text{which is impossible as} \\ \cos\theta\,\,\,\,\text{lies between}-1\,\,\text{and}\,\,1.$ 

Hence, the system of equations $\,\,x-y=2\pi/3\,\,$ and $\,\,\cos x+\cos y=\frac 32\,\,$ does not have any real solution.

Solve : 

$\,9.\, \tan x+\tan y=1\rightarrow(1), \\ x+y=\frac{\pi}{4}\rightarrow(2) \\ \therefore \tan\left(\frac{\pi}{4}-y\right)+\tan y=1 \\ \Rightarrow \frac{1-\tan y}{1+\tan y}+\tan y=1 \\ \Rightarrow (1-\tan y)+\tan y(1+\tan y)=1+\tan y \\ \Rightarrow 1-\tan y+\tan y+\tan^2y=1+\tan y \\ \Rightarrow \tan^2y-\tan y=0 \\ \Rightarrow \tan y(\tan y-1)=0 \\ \therefore \tan y=0 , \tan y=1 \rightarrow(1)$

Hence from $\,(1)\,$ we get, $\,\,\,\tan y=0 \\ \Rightarrow y=0 \\ \therefore x=\frac{\pi}{4}$

Again, by $\,(1)\,\,$ we get, $\,\tan y=1 \\ \Rightarrow y=\frac{\pi}{4} \\ \therefore x=0$

Hence follows the solution.