সৃজনশীলঃ ‘f(x) = sin…’ রাজশাহী বোর্ড ২০২১ – বিপরীত ত্রিকোনমিতিক ফাংশন ও ত্রিকোণমিতিক সমীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ২য় পত্র (এইচএসসি)

দেওয়া আছে,

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

$\therefore f\left(\pi f(\frac{\pi }{2}-\theta )\right)=f(\frac{\pi }{2}\pm \pi f(\theta \left)\right)$

$\Rightarrow f(\pi \sin (\frac{\pi }{2}-\theta ))=f(\frac{\pi }{2}\pm \pi \sin \theta )$

$\Rightarrow f(\pi \cos \theta )=\sin (\frac{\pi }{2}\pm \pi \sin \theta )$

$\Rightarrow \sin (\pi \cos \theta )=\sin (\frac{\pi }{2}\pm \pi \sin \theta )$

$\Rightarrow \pi \cos \theta =\frac{\pi }{2}\pm \pi \sin \theta$

$\Rightarrow \cos \theta =\frac{1}{2}\pm \sin \theta$

$\Rightarrow (\cos \theta \pm \sin \theta {)}^2={\left(\frac{1}{2}\right)}^2$

$\Rightarrow {\cos }^2\theta +{\sin }^2\theta \pm 2\sin \theta \cos \theta =\frac{1}{4}$

$\Rightarrow 1\pm \sin 2\theta =\frac{1}{4}$

$\Rightarrow \pm \sin 2\theta =\frac{1}{4}-1=-\frac{3}{4}$

$\Rightarrow \sin 2\theta =\pm \frac{3}{4}$

$\Rightarrow 2\theta ={\sin }^{-1}(\pm \frac{3}{4})=\pm {\sin }^{-1}\left(\frac{3}{4}\right)$

$\therefore \theta =\pm \frac{1}{2}{\sin }^{-1}\left(\frac{3}{4}\right)$ (দেখানো হলো)

দেওয়া আছে,

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

$\therefore 1+f(\frac{\pi }{2}-2x)+f(\frac{\pi }{2}-4x)+f(\frac{\pi }{2}-6x)=0$

$\Rightarrow 1+\sin (\frac{\pi }{2}-2x)+\sin (\frac{\pi }{2}-4x)+\sin (\frac{\pi }{2}-6x)=0$

$\Rightarrow 1+\cos 2x+\cos 4x+\cos 6x=0$

$\Rightarrow \cos 4x+\cos 2x+(1+\cos 6x)=0$

$\Rightarrow 2\cos 3x\cos x+2{\cos }^{2}3x=0$

$\Rightarrow 2\cos 3x(\cos x+\cos 3x)=0$

$\therefore \cos 3x=0$

$\Rightarrow 3x=(2n+1)\frac{\pi }{2}$

$\therefore x=(2n+1)\frac{\pi }{6},n\in Z$

অথবা, $ cos x+cos 3 x=0 $

$\Rightarrow 2\cos 2x\cos x=0$

$\therefore \cos 2x=0$

$\Rightarrow 2x=(2n+1)\frac{\pi }{2}$

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

অথবা, $\cos x=0$

$\therefore x=(2n+1)\frac{\pi }{2},n\in Z$

নির্ণের সমাধান: $x=(2n+1)\frac{\pi }{2},(2n+1)\frac{\pi }{4},(2n+1)\frac{\pi }{6}n\in Z$