Differentiation (Part-38) | S N De

 

Solve :

$\,1(i)\,\,\sin2\theta=\cos3\theta\,\,\left(0<\theta<\pi\right) \\ \Rightarrow \cos3\theta=\sin2\theta \\ \Rightarrow \cos3\theta=\cos(\frac{\pi}{2}-2\theta) \\ \Rightarrow 3\theta=2n\pi \pm \left(\frac{\pi}{2}-2\theta\right),\,\,n \in \mathbb Z \\ \therefore 3\theta=2n\pi+\frac{\pi}{2}-2\theta \rightarrow(1), \\ 3\theta=2n\pi-\frac{\pi}{2}+2\theta\rightarrow (2)$

From $\,(1)\,$ we get,  $\,\,5\theta=\frac{4n+1}{2} \pi \\ \Rightarrow \theta=\left(\frac{4n+1}{10}\right)\pi \rightarrow(3)$

Again, from $\,\,(2)\,\,$ we get, $\,\,\theta=\left(\frac{4n-1}{2}\right) \pi \rightarrow(4)$

Now, putting $\,\,n=0,1,2\,\,$ respectively in $\,(3),(4)\,\,$ we get,

$\,\theta=\frac{\pi}{10}\,\,\text{or}\quad \theta=-\pi/2\,\,\,\text{for}\,(n=0) \\ \theta=\frac{\pi}{2}\,\,\text{or}\quad\theta=3\pi/2\,\,\,\text{for}\,(n=1)\\\theta=\frac{9\pi}{10}\,\,\text{or}\quad \theta=7\pi/2\,\,\,\text{for}\,(n=2)$

Hence, $\,\,\theta=\frac{\pi}{10}, \frac{\pi}{2}, \frac{9\pi}{10}\,\,\left(0<\theta<\pi\right)$

$\,\,1(ii)\,\,\cos2\theta-\cos4\theta=0\,\,\,\left(0^{\circ}<\theta<360^{\circ}\right) \\ \Rightarrow 2 \sin3\theta\sin\theta=0\,\,[**] \\ \Rightarrow \sin3\theta=0 \rightarrow (1),\\ \sin\theta=0\longrightarrow(2)$

Now, from $\,(1)\,$ we get, $\,\,3\theta=n\pi \Rightarrow \theta=n\pi/3\rightarrow(3)$

Again, from $\,(2)\,$ we get, $\,\,\theta=n\pi\rightarrow(4)[n \in\mathbb Z]$

Since $\,\,0^{\circ}<\theta<360^{\circ},\,\,$ putting $\,\,n=0,1,2,3,4,5\,\,$ respectively, we get from $\,\,(3), (4)\,\,$

$\,\,\theta=\frac{\pi}{3}\,\,\text{or}\quad \theta=\pi\,\,\text{for}\,\,(n=1), \\ \theta=2\pi/3 \quad \text{for}\,\,(n=2)  \\ \theta=\pi \quad \text{for}\,\,(n=3)  \\ \theta=4\pi/3 \quad \text{for}\,\,(n=4)  \\ \theta=5\pi/3 \quad \text{for}\,\,(n=5) $ 

Hence, $\,\,\theta=\frac{\pi}{3},\frac{2\pi}{3},\pi,\frac{4\pi}{3},\frac{5\pi}{3}\,\,\text{i.e.}\\ \theta= 60^{\circ},120^{\circ},180^{\circ},240^{\circ},300^{\circ}.$

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

$\,1(iii)\,\,(2\sin^2x-1)+1=0 \\ \Rightarrow 1-(1-2\sin^2x)=0 \\ \Rightarrow 1-\cos2x=0 \\ \Rightarrow \cos2x=1 \\ \Rightarrow 2x=2n\pi,\,\,n\in\mathbb Z \\ \Rightarrow x=n\pi\,\,\text{(ans.)}$

$\,\,1(iv)\,\,\cos m\theta -\sin n\theta =0 \\ \Rightarrow \cos m\theta =\sin n\theta \\ \Rightarrow \cos m\theta =\cos\left(\pi/2-n\theta \right) \\ \Rightarrow m\theta =2p\pi \pm \left(\pi/2-n\theta \right) \rightarrow(1)$

Hence, from $\,\,(1)\,\,$ we get, $\,\,m\theta =\left(\frac{4p+1}{2}\right)\pi-n\theta \rightarrow(2) \\ \text{or}\,\,\,m\theta =\left(\frac{4p-1}{2}\right)\pi+n\theta \rightarrow(3)$

So, from $\,\,(2)\,\,$ we get, $\,(m+n)\theta=\left(\frac{4p+1}{2}\right)\pi \\ \Rightarrow \theta=\left(\frac{4p+1}{2(m+n)}\right)\pi$

Again , from $\,(3)\,$ we get, $\,(m-n)\theta=\left(\frac{4p-1}{2}\right)\pi \\ \Rightarrow \theta=\left(\frac{4p-1}{2(m-n)}\right)\pi,\,\,p\in\mathbb  Z$

$\,1(v)\,\, \cos m\theta+\cos n\theta=0 \\ \Rightarrow \cos m\theta=-\cos n\theta \\ \Rightarrow \cos m\theta=\cos(\pi- n\theta) \\ \Rightarrow m\theta=2p\pi \pm (\pi-n\theta),\,\,p \in \mathbb Z\rightarrow(1)$

Hence, from $\,(1)\,$ , we get, $\,\,m\theta=2p\pi+\pi-n\theta \\ \Rightarrow (m+n)\theta=(2p+1)\pi \\ \Rightarrow \theta=\frac{(2p+1)\pi}{m+n} \rightarrow(2)$

Again, by $\,(1)\,$ we get, $\,m\theta=2p\pi-\pi+n\theta \\ \Rightarrow (m-n)\theta=(2p-1)\pi \\ \therefore \theta=\frac{(2p-1)\pi}{m-n}\rightarrow(3)$

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

To continue with GENERAL SOLUTIONS OF TRIGONOMETRIC EQUATIONS (PART-6), click here.