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GENERAL SOLUTIONS OF TRIGONOMETRIC EQUATIONS (PART-2)

GENERAL SOLUTIONS OF TRIGONOMETRIC EQUATIONS (PART-2)


 $\,\,1(xi)\,\, 1+\sin^2\theta=3\sin\theta\cos\theta,\,\,$ given that , $\,\,\cot18^{\circ}26'=3.$

Sol. $\,\, 1+\sin^2\theta=3\sin\theta\cos\theta \\ \Rightarrow 2+2\sin^2\theta=3. 2\sin\theta\cos\theta \\ \Rightarrow 2+1-\cos2\theta=3\sin2\theta \\ \Rightarrow 3\sin2\theta+\cos2\theta=3 \\ \Rightarrow \frac{3}{\sqrt{3^2+1^2}}\sin2\theta+\frac{1}{\sqrt{3^2+1^2}}\cos2\theta=\frac{3}{\sqrt{3^2+1^2}} \\ \Rightarrow \frac{3}{\sqrt{10}}\sin2\theta+\frac{1}{\sqrt{10}}\cos2\theta=\frac{3}{\sqrt{10}}\rightarrow(1)$

Let $\,\,\alpha=71^{\circ}34',\,\,$ 

so that $\,\,\tan \alpha=\cot(90^{\circ}-\alpha)=\cot18^{\circ}26'=3.$

Now, $\,\,\tan\alpha=3 \\ \Rightarrow \frac{\sin\alpha}{\cos\alpha}=\frac 31 \\ \Rightarrow \frac{\sin\alpha}{3}=\frac{\cos\alpha}{1}=\frac{\sqrt{\sin^2\alpha+\cos^2\alpha}}{\sqrt{3^2+1^2}}=\frac{1}{\sqrt{10}} \\ \Rightarrow \sin\alpha=\frac{3}{\sqrt{10}},\\ \text{and}\,\,\cos\alpha=\frac{1}{\sqrt{10}}.$

Hence, from $\,(1)\,$ we get, 

$\,\,\sin2\theta\sin\alpha+\cos2\theta\cos\alpha=\sin\alpha \\ \Rightarrow \cos(2\theta-\alpha)=\sin\alpha=\cos(\frac{\pi}{2}-\alpha) \\ \Rightarrow 2\theta-\alpha=2n\pi \pm (\frac{\pi}{2}-\alpha)\rightarrow(2)$

Hence, from $\,(2)\,$ we get, $\,\,2\theta-\alpha=2n\pi+\pi/2-\alpha \\ \Rightarrow \theta=n\pi+\pi/4.$ 

Again, from $\,(2)\,$, we get, $\,\,2\theta-\alpha=2n\pi-\pi/2+\alpha \\ \Rightarrow 2\theta=2n\pi+2\alpha-\pi/2 \\ \Rightarrow \theta=n\pi+\alpha-\pi/4\\~~~~~~~=n\pi+71^{\circ}34'-45^{\circ}\\~~~~~~~=n\pi+26^{\circ}34'.$

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

$\,\,1(xii)\,\,\tan\theta+\sec\theta=\sqrt3 \\ \Rightarrow \frac{\sin\theta}{\cos\theta}+\frac{1}{\cos\theta}=\sqrt3 \\ \Rightarrow \sin\theta+1=\sqrt3\cos\theta \\ \Rightarrow \sqrt3\cos\theta-\sin\theta=1 \\ \Rightarrow \frac{\sqrt3}{2}\cos\theta-\frac 12\sin\theta=\frac 12 \\ \Rightarrow \cos\frac{\pi}{6}\cos\theta-\sin\frac{\pi}{6}\sin\theta=\cos\frac{\pi}{3} \\ \Rightarrow \cos(\theta+\frac{\pi}{6})=\cos\frac{\pi}{3} \\ \Rightarrow \theta+\frac{\pi}{6}=2n\pi\pm \pi/3 \rightarrow(1)$

So, from $\,(1)\,$ we get, $\,\theta=2n\pi+\pi/3-\pi/6=2n\pi+\pi/6\rightarrow(2), \\ \theta=2n\pi-\pi/3-\pi/6=2n\pi-\pi/2\rightarrow(3).$

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

$\,\,1(xiii)\,\,(1-\tan\theta)(1+\sin2\theta)=1+\tan\theta \\ \Rightarrow (1-\tan\theta) \left[1+\frac{2\tan\theta}{1+\tan^2\theta}\right]=1+\tan\theta \\ \Rightarrow (1-\tan\theta)\left[\frac{1+\tan^2\theta+2\tan\theta}{1+\tan^2\theta}\right]-(1+\tan\theta)\\=0 \\ \Rightarrow (1-\tan\theta)\frac{(1+\tan\theta)^2}{1+\tan^2\theta}-(1+\tan\theta)=0 \\ \Rightarrow (1+\tan\theta)\left[\frac{(1-\tan\theta)(1+\tan\theta)}{1+\tan^2\theta}-1\right]=0 \\ \Rightarrow (1+\tan\theta)\left[\frac{1-\tan^2\theta-1-\tan^2\theta}{1+\tan^2\theta}\right]=0 \\ \Rightarrow (1+\tan\theta)\times (-2\tan^2\theta)=0 \rightarrow(1) \\ \therefore 1+\tan\theta=0 \\ \Rightarrow \tan\theta=-1 \\ \Rightarrow \theta=n\pi+\frac{3\pi}{4}\rightarrow(2)$

Again , from $\,\,(1)\,\,$, we get , $\tan\theta=0 \Rightarrow \theta=n\pi\rightarrow(3).$ 

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

$\,\,1(xiv)\,\,\sec\theta+1=(2+\sqrt3)\tan\theta  \,\,[0<\theta<2\pi]\\ \Rightarrow \frac{1}{\cos\theta}+1=(2+\sqrt3)\frac{\sin\theta}{\cos\theta} \\ \Rightarrow (1+\cos\theta)=(2+\sqrt3)\sin\theta \\ \Rightarrow 2\cos^2\frac{\theta}{2}=(2+\sqrt3)\times 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} \\ \Rightarrow 2\cos\frac{\theta}{2}\left[\cos\frac{\theta}{2}-(2+\sqrt3)\sin\frac{\theta}{2}\right]=0 \rightarrow(1)$

Hence, from $\,(1)\,$  we get, $\,\,\cos\frac{\theta}{2}=0 \\ \Rightarrow \frac{\theta}{2}=(2n+1)\frac{\pi}{2}\\ \Rightarrow \theta=(2n+1)\pi \rightarrow(2) $ 

Again, from $\,\,(1)\,\,$ we get, $\,\,\cos\frac{\theta}{2}=(2+\sqrt3)\sin\frac{\theta}{2} \\ \Rightarrow \tan\frac{\theta}{2}=\frac{1}{2+\sqrt3}=2-\sqrt3 \\ \Rightarrow \tan\frac{\theta}{2}=\tan\frac{\pi}{12} \\ \Rightarrow \frac{\theta}{2}=n\pi+\frac{\pi}{12} \\ \Rightarrow \theta=2n\pi+\frac{\pi}{6}\rightarrow(3)$

Now, putting $\,\,n=0\,\,$ in $\,\,(2),(3)\,\,$ we get, $\,\,\theta=\pi,\frac{\pi}{6}\,\,\text{since},\,\,0<\theta<2\pi.$ 

$\,1(xv)\,\,\cot\theta+\cot(\frac{\pi}{4}+\theta)=2 \\ \Rightarrow \cot\theta+\frac{\cot\frac{\pi}{4}\cot\theta-1}{\cot\frac{\pi}{4}+\cot\theta}=2 \\ \Rightarrow \cot\theta+\frac{\cot\theta-1}{1+\cot\theta}=2 \\ \Rightarrow \cot\theta(1+\cot \theta)+\cot\theta-1=2(1+\cot\theta) \\ \Rightarrow \cot\theta+\cot^2\theta+\cot\theta-1=2+2\cot\theta \\ \Rightarrow \cot^2\theta=3 \\ \Rightarrow \tan^2\theta=\frac 13 \\ \Rightarrow \tan\theta=\pm\frac{1}{\sqrt3}=\tan(\pm\frac{\pi}{6}) \\ \therefore \theta=n\pi \pm \frac{\pi}{6}\,\,\text{(ans.)}$ 

$\,1(xvi)\,\,\cot^3\theta+6\csc2\theta-8\csc^32\theta=0 \\ \Rightarrow \frac{\cos^3\theta}{\sin^3\theta}+\frac{6}{2\sin\theta\cos\theta}-\frac{8}{8\sin^3\theta\cos^3\theta}=0  \\ \Rightarrow \cos^6\theta+3\sin^2\theta\cos^2\theta-1=0\\ \Rightarrow \cos^6\theta+3(1-\cos^2\theta)\cos^2\theta-1=0 \\ \Rightarrow \cos^6\theta+3\cos^2\theta-3\cos^4\theta-1=0\\ \Rightarrow \cos^6\theta-3\cos^4\theta+3\cos^2\theta-1=0 \\ \Rightarrow (\cos^2\theta-1)^3=0 \\ \Rightarrow \cos\theta=\pm1\rightarrow(1)$

But this value of $\,\,\cos\theta\,$ from $\,(1)\,$ does not satisfy the equation. So, the given equation does not have any solution.

$\,\,1(xvii)\,\,\tan^22x+\cot^22x\\+2\tan2x+2\cot2x=6 \\ \Rightarrow (\tan2x+\cot2x)^2\\+2(\tan2x+\cot2x)+1=6+2+1 \\ \Rightarrow (\tan2x+\cot2x+1)^2=3^2 \\ \Rightarrow \tan2x+\cot2x+1=\pm3 \rightarrow(1) $

Now, from $\,\,(1)\,$ we get ,

$\tan2x+\cot2x+1=3 \\ \Rightarrow\,\,(\sqrt{\tan2x})^2-2\sqrt{\tan2x.\cot2x}\\+(\sqrt{\cot2x})^2=0 \\ \Rightarrow (\sqrt{\tan2x}-\sqrt{\cot2x})^2=0 \\ \Rightarrow \sqrt{\tan2x}-\sqrt{\cot2x}=0 \\ \therefore \tan2x=1=\tan\frac{\pi}{4} \\ \Rightarrow 2x=n\pi+\frac{\pi}{4}\\ \Rightarrow x=\frac{n\pi}{2}+\frac{\pi}{8}\\ \Rightarrow x=(4n+1)\frac{\pi}{8}\rightarrow(2),\\  n\,\,\,\text{being any integer.}$

Again, by $\,(1)\,$ we get,

$\,\,\tan2x+\cot2x+1=-3 \\ \Rightarrow \frac{\sin2x}{\cos2x}+\frac{\cos2x}{\sin2x}=-3-1\\ \Rightarrow \frac{\sin^22x+\cos^22x}{\sin2x\cos2x}=-4 \\ \Rightarrow \frac{1}{\sin2x\cos2x}=-4 \\ \Rightarrow -4\sin2x\cos2x=1 \\ \Rightarrow 2\sin2x\cos2x=-\frac 12 \\ \Rightarrow \sin4x=-\frac 12=\sin\left(-\frac{\pi}{6}\right) \\ \Rightarrow 4x=n\pi+(-1)^n\left(-\frac{\pi}{6}\right) \\ \Rightarrow x=\frac{n\pi}{4}-(-1)^n\frac{\pi}{24}\rightarrow(3)$

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

$\,1(xviii)\,\,\sin^3x+\sin^3\left(x-\frac{2\pi}{3}\right)\\+\sin^3\left(x+\frac{2\pi}{3}\right)=0\\ \Rightarrow 4\sin^3x+4\sin^3\left(x-\frac{2\pi}{3}\right)\\+4\sin^3\left(x+\frac{2\pi}{3}\right)=0 \\ \Rightarrow 3\left[\sin x+\sin\left(x-\frac{2\pi}{3}\right)\\+\sin\left(x+\frac{2\pi}{3}\right)\right]-3\sin3x=0\,\,\,[**]\\ \Rightarrow 3\left[\sin x+ 2\sin x\cos \frac{2\pi}{3}\right]\\-3\sin3x=0\,\,[*] \\ \Rightarrow 3\left[\sin x+2\sin x\times \left( \frac {-1}{2}\right) \right]\\-3\sin 3x=0 \\ \Rightarrow 3\left[\sin x-\sin x\right]-3\sin 3x=0 \\ \therefore \sin 3x=0=\sin n\pi \\ \therefore x=\frac{n\pi}{3}$

Note[**] : $\,\,4\sin^3x=3\sin x-\sin 3x $

Note[*] : $\,\,\sin(A+B)+\sin(A-B)=2\sin A\cos B$

$\,\,1(xix)\,\,2\tan x=\sin 4x-2\sin2x\cos2x \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\times \tan^2x \\ \Rightarrow 2\tan x=2\sin2x\cos2x(1-\tan^2x) \\ \Rightarrow \frac{2\tan x}{1-\tan^2x}=2\sin2x\cos2x \\ \Rightarrow \tan2x=2\sin2x\cos2x \\ \Rightarrow \frac{\sin2x}{\cos2x}=2\sin2x\cos2x \\ \Rightarrow  \sin2x=2\sin2x\cos^22x \\ \Rightarrow \sin2x\left(1-2\cos^22x\right)=0 \\ \therefore \sin 2x=0 \rightarrow(1),\\ 1-2\cos^22x=0\rightarrow(2)$

Now, from $\,(1)\,$ we get, $\,\sin 2x=0 \Rightarrow 2x=n\pi \Rightarrow x=\frac{n\pi}{2}$

Again. from $\,(2)\,$ we get, $\,\,\cos2x=\frac{1}{\sqrt2}=\cos\frac{\pi}{4} \\ \Rightarrow 2x=2n\pi+\pi/4 \\ \Rightarrow x=n\pi+\pi/8.$

Also, from $\,(2)\,$ we get, $\,\cos2x=-\frac{1}{\sqrt2}=\cos\frac{3\pi}{4} \\ \Rightarrow 2x=2n\pi+\frac{3\pi}{4} \\ \Rightarrow x=n\pi+\frac{3\pi}{8},\,\,\,n \,\,\text{being any integer.}$ 

$\,\,1(xx)\,\,\sin3\alpha=4\sin\alpha\sin(x+\alpha)\sin(x-\alpha) \\ \Rightarrow \sin 3\alpha=4\sin\alpha(\sin^2x-\sin^2\alpha) \\ \Rightarrow 3\sin\alpha-4\sin^3\alpha=4\sin\alpha\sin^2x-4\sin^3\alpha \\ \Rightarrow 3\sin\alpha=4\sin\alpha\sin^2x \\ \therefore 3=4\sin^2x \\ \Rightarrow \sin x=\pm \frac{\sqrt3}{2}=\sin\left(\pm \frac{\pi}{3}\right) \\ \Rightarrow x=n\pi \pm\frac{\pi}{3}\,\,\text{(ans.)}$

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