সমাধান কর : $\sin (x) + \sin (\frac{x}{2}) = 0,$ যখন $0 \leq x \leq 2 π$

Created: 1 year ago | Updated: 1 year ago

Updated: 1 year ago

$\sin x + \sin \frac{x}{2} = 0 \Rightarrow 2 \sin \frac{x}{2} \cos \frac{x}{2} \sin \frac{x}{2} = 0$

$\Rightarrow \sin \frac{x}{2} (2 \cos \frac{x}{2} + 1) = 0$

∴ হয় $\sin \frac{x}{2} = 0 \Rightarrow \frac{x}{2} = n π \Rightarrow x = 2 n π$

$0 \leq x \leq 2 π$ ব্যবধিতে, n=0, 1 ∴ x =0, $2 π$

অথবা, $2 \cos \frac{x}{2} + 1 = 0 \Rightarrow \cos \frac{x}{2} = - \frac{1}{2} = \cos \frac{2 π}{3}$

$\Rightarrow \frac{x}{2} = 2 n π \pm \frac{2 π}{3} ∴ x = 4 n π \pm \frac{4 π}{3}$

∴ $0 \leq x \leq 2 π$ ব্যবধিতে, $n = 0 ∴ x = \frac{4 π}{3}$

∴ নির্ণেয় সমাধান : $x = 0, \frac{4 π}{3}, 2 π$

1 year ago

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