সমীকরণ সমাধান
cosx+3sinx=2\cos{x}+\sqrt3\sin{x}=\sqrt2 cosx+3sinx=2 হলে, x- এর মান কত?
(24n+7)π12, (24n+5)π12\left(24n+7\right)\frac{\pi}{12},\ \left(24n+5\right)\frac{\pi}{12}(24n+7)12π, (24n+5)12π
(24n+3)π12 , (12n+1)π12\left(24n+3\right)\frac{\pi}{12}\ ,\ \left(12n+1\right)\frac{\pi}{12}(24n+3)12π , (12n+1)12π
(24n+1)π12, (24n+7)π12\left(24n+1\right)\frac{\pi}{12},\ \left(24n+7\right)\frac{\pi}{12}(24n+1)12π, (24n+7)12π
(24n+3)π12 , (24n+5)π12\left(24n+3\right)\frac{\pi}{12}\ ,\ \ \left(24n+5\right)\frac{\pi}{12}(24n+3)12π , (24n+5)12π
12cosx+32sinx=12⇒cos(x−π3)=cos(π4)⇒x−π3=2nπ±π4⇒x=2nπ+π12, 2nπ+7π12\frac{1}{2}\cos{x}+\frac{\sqrt3}{2}\sin{x}=\frac{1}{\sqrt2}\Rightarrow\cos{\left(x-\frac{\pi}{3}\right)}=\cos{\left(\frac{\pi}{4}\right)}\Rightarrow x-\frac{\pi}{3}=2n\pi\pm\frac{\pi}{4} \Rightarrow x=2n\pi+\frac{\pi}{12},\ 2n\pi+\frac{7\pi}{12}21cosx+23sinx=21⇒cos(x−3π)=cos(4π)⇒x−3π=2nπ±4π⇒x=2nπ+12π, 2nπ+127π
⇒x=(24n+1)π12, (24n+7)π12\Rightarrow x=\left(24n+1\right)\frac{\pi}{12},\ \left(24n+7\right)\frac{\pi}{12}⇒x=(24n+1)12π, (24n+7)12π
f(x)=sinx \mathrm{f}(x)=\sin x f(x)=sinx এবং g(x)=cosx g(x)=\cos x g(x)=cosx.
sin−1x\sin^{-1}xsin−1x এর মূখ্যমানের সীমা নিচের কোনটি?
3sec-1(2)=cos-1x হলে x এর মান ত?
cosθ=12 \cos{θ} = \frac{1}{\sqrt{2}} cosθ=21 হলে,θ এর মান কত?