সমীকরণ সমাধান
সমাধান কর: 4(sin2θ+cosθ)=5;−π<θ<π4(\sin^2θ+\cosθ)=5;−\pi<θ<\pi4(sin2θ+cosθ)=5;−π<θ<π
π/3
-π/3
Both A & B
None
4(sin2θ+cosθ)=5⇒4(1−cos2θ+cosθ)−5=0⇒(2cosθ−1)2=0⇒θ=±60∘=±π3 \begin{array}{l}\quad 4\left(\sin ^{2} \theta+\cos \theta\right)=5 \\ \Rightarrow 4\left(1-\cos ^{2} \theta+\cos \theta\right)-5=0 \Rightarrow(2 \cos \theta-1)^{2}=0 \\ \Rightarrow \theta= \pm 60^{\circ}= \pm \frac{\pi}{3}\end{array} 4(sin2θ+cosθ)=5⇒4(1−cos2θ+cosθ)−5=0⇒(2cosθ−1)2=0⇒θ=±60∘=±3π
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(abtanθ2)=?2 \tan ^{-1}\left(\sqrt{\frac{a}{b}} \tan \frac{\theta}{2}\right) = ? 2tan−1(batan2θ)=?
