সৃজনশীলঃ ‘f(x)=cosecx−cot⁡x,g(x)=sin⁡x…’ ময়মনসিংহ বোর্ড ২০২১ – বিপরীত ত্রিকোনমিতিক ফাংশন ও ত্রিকোণমিতিক সমীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ২য় পত্র (এইচএসসি)

দেওয়া আছে,

$f\left(x\right)=\mathrm{cosec}x-\cot x$

$\Rightarrow f\left(\theta \right)=\mathrm{cosec}\theta -\cot \theta$

$\Rightarrow \mathrm{cosec}\theta -\cot \theta =\frac{3}{4}$

$\Rightarrow \frac{1-\cos \theta }{\sin \theta }=\frac{3}{4}\Rightarrow \frac{(1-\cos \theta {)}^2}{{\sin }^2\theta }=\frac{9}{16}$

$\Rightarrow 16(1-2\cos \theta +{\cos }^2\theta )=9{\sin }^2\theta$

$\Rightarrow 16-32\cos \theta +16{\cos }^2\theta =9-9{\cos }^2\theta$

$\Rightarrow 25{\cos }^2\theta -25\cos \theta -7\cos \theta +7=0$

$\Rightarrow 25\cos \theta (\cos \theta -1)-7(\cos \theta -1)=0$

$\Rightarrow (\cos \theta -1)(25\cos \theta -7)=0$

$\therefore \cos \theta =1$

অথবা,

$25\cos \theta =7$ [গ্রহনযোগ্য নয়]

$\Rightarrow \cos \theta =\frac{7}{25}$

$\Rightarrow \sin \theta =\pm \frac{24}{25}$

$\Rightarrow \theta ={\sin }^{-1}(\pm \frac{24}{25})$

$\therefore \theta =\pm {\sin }^{-1}\left(\frac{24}{25}\right)$

(দেখানো হল)

দেওয়া আছে,

$g\left(x\right)=\sin x$

এখন, $g\left(5\theta \right)-\sqrt{3}g\left(\theta \right)=g\left(3\theta \right)$

$\Rightarrow \sin 5\theta -\sqrt{3}\sin \theta =\sin 3\theta$

$\Rightarrow \sin 5\theta -\sin 3\theta =\sqrt{3}\sin \theta$

$\Rightarrow 2\cos 4\theta \sin \theta -\sqrt{3}\sin \theta =0$

$\Rightarrow \sin \theta (2\cos 4\theta -\sqrt{3})=0$

$\therefore \sin \theta =0$

$\Rightarrow \theta =n\pi$. n€Z $

$or,2cos4\theta =\sqrt{3}$

$\Rightarrow cos4\theta$

$=\frac{\sqrt{3}}{2}=cos\frac{\pi }{6}$

$\Rightarrow 4\theta =2n\pi \pm \frac{\pi }{6}$

$\Rightarrow \theta =\frac{n\pi }{2}\pm \frac{\pi }{24}nϵz$

$\therefore$ নির্ণেয় সমাধান : $\theta =\frac{n\pi }{2}\pm \frac{\pi }{24}nϵz$