সমীকরণ সমাধান
if φ(z)=ysinz+v এবং Φ(w)=sin-1(yw2+y2)-1 হয়, তবে φ(Φ(u2)) এর মান হলো -
(u4+y)-1+v
ysin-1(u2+y)+v
ysiny(u4+y)+v
(u4+y)-2+v
ϕ(ψ(u2)=ϕ[sin−1(yu4+y2)−1]=ysin[sin−1(yu4+y2)−1]+v=y×1yu4+y2+v=1u4+y+v=(u4+y)−1+v \begin{aligned} \phi(\psi(u^2) & =\phi\left[\sin ^{-1}\left(y u^{4}+y^{2})^-1\right]\right. \\ & =y \sin \left[\sin ^{-1}\left(y u^{4}+y^{2}\right)^{-1}\right]+v \\ & =y \times \frac{1}{y u^{4}+y^{2}}+v \\ & =\frac{1}{u^{4}+y}+v \\ & =\left(u^{4}+y\right)^{-1}+v\end{aligned} ϕ(ψ(u2)=ϕ[sin−1(yu4+y2)−1]=ysin[sin−1(yu4+y2)−1]+v=y×yu4+y21+v=u4+y1+v=(u4+y)−1+v
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?
সমাধান কর: 4(sin2θ+cosθ)=5;−π<θ<π4(\sin^2θ+\cosθ)=5;−\pi<θ<\pi4(sin2θ+cosθ)=5;−π<θ<π
