সমীকরণ সমাধান
sin θ + 1 =0 হলে θ =?
(4n−1)π2,n∈Z \left ( 4 n - 1 \right ) \frac{\pi}{2} , n \in \mathbb{Z} (4n−1)2π,n∈Z
(4n+1)π2,n∈Z \left ( 4 n + 1 \right ) \frac{\pi}{2} , n \in \mathbb{Z} (4n+1)2π,n∈Z
(2n+1)π2,n∈Z \left ( 2 n + 1 \right ) \frac{\pi}{2} , n \in \mathbb{Z} (2n+1)2π,n∈Z
(2n−1)π2,n∈Z \left ( 2 n - 1 \right ) \frac{\pi}{2} , n \in \mathbb{Z} (2n−1)2π,n∈Z
.sinθ+1=0⇒sinθ=−1θ=(4n− 1)π2,n∈Z \begin{array}{l}\text { }^{} . \sin \theta+1=0 \Rightarrow \sin ^{} \theta=-1 \\ \theta=(4 n- \ 1) \frac{\pi}{2}, n \in \mathbb{Z} \text { }\end{array} .sinθ+1=0⇒sinθ=−1θ=(4n− 1)2π,n∈Z
f(x)=sinx \mathrm{f}(x)=\sin x f(x)=sinx এবং g(x)=cosx g(x)=\cos x g(x)=cosx.
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?
2tan−1(cosx)=tan−1(2cosecx) 2 \tan^{- 1}{\left ( \cos{x} \right )} = \tan^{- 1}{\left ( 2 \cos{e} c x \right )} 2tan−1(cosx)=tan−1(2cosecx) এর সমাধান -
