Differentiation (Part-38) | S N De

 

$\,1(vii)\,\, \sin2x+\sin4x+\sin6x=0 \\ \Rightarrow (\sin 2x+\sin 6x)+\sin4x=0 \\ \Rightarrow2\sin\left(\frac{6x+2x}{2}\right)\cos\left(\frac{6x-2x}{2}\right) +\sin4x=0 \\ \Rightarrow2\sin4x\cos2x+\sin4x=0 \\ \Rightarrow\sin4x(2\cos2x+1)=0\\ \Rightarrow\sin4x=0\rightarrow(1),\\ \text{or,}\,\,2\cos2x+1=0\rightarrow(2).$

From $\,(1)\,$ we get, $\,\,4x=n\pi,\,\,n\in\mathbb Z \\ \Rightarrow x=\frac{n\pi}{4}.$

By $\,(2)\,\,$ we get, $\,\,\cos2x=-\frac 12=\cos\left(\frac{2\pi}{3}\right)\\ \Rightarrow2x=2n\pi \pm\frac{2\pi}{3},\,\,n\in\mathbb Z \\ \Rightarrow x=n\pi \pm\frac{\pi}{3}.$

$\,\,\therefore \,\,x=\frac{n\pi}{4},\,\,\,n\pi \pm\frac{\pi}{3},\,\,n\in\mathbb Z.$

$\,1(viii)\,\,\cos x-\sin3x=\cos2x \\ \Rightarrow \cos x-\cos 2x-\sin3x=0\\ \Rightarrow 2\sin\frac{3x}{2}\sin\frac x2-2\sin\frac{3x}{2}\cos\frac{3x}{2}=0\,\,[**]\\ \Rightarrow2\sin\frac{3x}{2}\left(\sin\frac x2-\cos\frac{3x}{2}\right) =0 \\ \Rightarrow 2\sin\frac{3x}{2}=0\rightarrow(1),\\ \text{or,}\,\,\sin\frac x2-\cos\frac{3x}{2}=0\rightarrow(2)$

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

By $\,(2)\,\,$ we get, $\,\,\cos\frac{3x}{2}=\cos\left(\frac{\pi}{2}-\frac x2\right) \\ \Rightarrow\frac{3x}{2}=2n\pi \pm \left(\pi/2-x/2\right) \\ \Rightarrow x=(4n+1)\frac{\pi}{4}, \,\,(4n-1)\frac{\pi}{2},\, n\in\mathbb Z.$

$\,\therefore \,\,\, x=\frac{2n\pi}{3},\,\,(4n+1)\frac{\pi}{4},\,\,(4n-1)\frac{\pi}{2},\,\,n \in\mathbb Z.$

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

$\,1(ix)\,\,\sin\theta+\sin2\theta+\sin3\theta+\sin4\theta=0 \\ \Rightarrow (\sin\theta+\sin3\theta)+(\sin2\theta+\sin4\theta)=0 \\ \Rightarrow 2\sin2\theta\cos\theta+2\sin3\theta\cos\theta=0\,\,[*] \\ \Rightarrow 2\cos\theta(\sin2\theta+\sin3\theta)=0 \\ \Rightarrow 4\cos\theta\sin\frac{5\theta}{2}\cos\frac{\theta}{2}=0 \\ \Rightarrow \cos \theta=0\rightarrow(1), \\ \text{or,}\,\, \sin\frac{5\theta}{2}=0\rightarrow(2) \\ \text{or,}\,\,\cos\frac{\theta}{2}=0\rightarrow(3).$

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

By $\,(2)\,\,$, we get, $\,\,\frac{5\theta}{2}=n\pi \Rightarrow \theta=\frac{2n\pi}{5},\,\,n\in\mathbb Z$

Finally, $\,(3)\,\,$ gives $\,\,\frac{\theta}{2}=(2n+1)\frac{\pi}{2}\\ \Rightarrow \theta=(2n+1)\pi,\,\,n\in\mathbb Z.$

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

$\,1(x)\,\,\sin3x+\sin x=\cos6x+\cos4x \\ \Rightarrow 2\sin2x\cos x=2\cos5x\cos x \\ \Rightarrow 2\cos x(\sin2x-\cos5x)=0 \\ \Rightarrow \cos x=0\rightarrow(1),\\ \text{or,}\,\,\sin2x-\cos5x=0\rightarrow(2).$

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

By $\,(2)\,\,$ we get, $\,\,\cos 5x=\sin 2x \\ \Rightarrow\cos5x=\cos(\pi/2-2x) \\ \Rightarrow 5x=2n\pi \pm \left(\frac{\pi}{2}-2x\right) \\ \Rightarrow x=(4n+1)\frac{\pi}{14},\,\,(4n-1)\frac{\pi}{6}, \\ \text{where}\,\,n\in\mathbb Z \rightarrow(4)$

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

$\,1(xi)\,\,\sin5x-\sin3x-\sin x=0\, \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\left(0^{\circ}<x<360^{\circ}\right) \\ \Rightarrow 2\cos4x \sin x-\sin x=0 \\ \Rightarrow \sin x(2\cos 4x-1)=0 \\ \Rightarrow \sin x=0 \Rightarrow x=n\pi \rightarrow(1) \\ \text{or,}\,\,\,2\cos4x-1=0 \\ \Rightarrow \cos 4x=\frac 12=\cos\left(\frac{\pi}{3}\right) \\ \Rightarrow 4x=2n\pi \pm \frac{\pi}{3} \\ \Rightarrow x=\frac{n\pi}{2} \pm\frac{\pi}{12}\rightarrow(2)$

By $\,(1)\,\,$ we get, $\,\,x=\pi,\,\,\text{for}\,\,(n=1).$

By $\,\,(2)\,\,$ we get, $\,x= \pm \frac{\pi}{12}=\pm 15^{\circ}\,\,(n=0)\\ x=\frac{7\pi}{12}, \,\frac{5\pi}{12}=105^{\circ},75^{\circ} \,\,(n=1) \\ x=\frac{13\pi}{12},\frac{11\pi}{12}=195^{\circ},165^{\circ}\,\,(n=2) \\ x=\frac{19\pi}{12},\frac{17\pi}{12}=285^{\circ},255^{\circ}\,\,(n=3) \\ x=\frac{25\pi}{12},\frac{23\pi}{12}=375^{\circ},345^{\circ}\,\,(n=4)$

Since      $\,\,0^{\circ}<x<360^{\circ},\,$ we have,  $\,x=15^{\circ},75^{\circ},105^{\circ},165^{\circ},180^{\circ},195^{\circ},255^{\circ},\\ 285^{\circ},\,345^{\circ}.$

$\,1(xii)\,\,\cos 4\theta=\cos3\theta-\cos2\theta \\ \Rightarrow \cos4\theta+\cos2\theta-\cos3\theta=0\\ \Rightarrow 2\cos3\theta\cos\theta-\cos3\theta=0 \\ \Rightarrow \cos3\theta(2\cos\theta-1)=0 \\ \therefore \cos3\theta=0\rightarrow(1), \\ \text{or,}\,\,2\cos\theta-1=0\rightarrow(2)$

By $\,(1)\,$, we get, $\,\,3\theta=(2n+1)\frac{\pi}{2} \\ \Rightarrow \theta=(2n+1)\frac{\pi}{6} ,\,\,\,n \in \mathbb Z.$

Again by $\,(2)\,$, we get, $\,\,\cos\theta=\frac 12=\cos\frac{\pi}{3} \\ \Rightarrow \theta=2n\pi \pm\pi/3,\,n\in\mathbb Z. \\ \therefore \theta=(2n+1)\pi/6,\,2n\pi \pm \pi/3,\,\, n \in\mathbb Z.$

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