সমীকরণ সমাধান
সমাধান করঃ tan 2θ tanθ= 1, 0≤θ≤π/2
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tan2θtanθ=1⇒sin2θsinθcos2θcosθ=1⇒cos2θcosθ=sin2θsinθ⇒cos2θcosθ−sin2θsinθ=0⇒cos(2θ+θ)=0⇒cos3θ=0⇒3θ=cos−1(0)=90∘∴θ=30∘. \begin{aligned} & \tan 2 \theta \tan \theta=1 \\ \Rightarrow & \frac{\sin 2 \theta \sin \theta}{\cos 2 \theta \cos \theta}=1 \\ \Rightarrow & \cos 2 \theta \cos \theta=\sin 2 \theta \sin \theta \\ \Rightarrow & \cos 2 \theta \cos \theta-\sin 2 \theta \sin \theta=0 \\ \Rightarrow & \cos (2 \theta+\theta)=0 \\ \Rightarrow & \cos 3 \theta=0 \\ \Rightarrow & 3 \theta=\cos ^{-1}(0)=90^{\circ} \\ \therefore & \theta=30^{\circ} .\end{aligned} ⇒⇒⇒⇒⇒⇒∴tan2θtanθ=1cos2θcosθsin2θsinθ=1cos2θcosθ=sin2θsinθcos2θcosθ−sin2θsinθ=0cos(2θ+θ)=0cos3θ=03θ=cos−1(0)=90∘θ=30∘.
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
