সমীকরণ সমাধান
2cos2θ+22sinθ=3 2 \cos^{2}{\theta} + 2 \sqrt{2} \sin{\theta} = 3 2cos2θ+22sinθ=3 হলে, θ \theta θ এর মান --
30°
45°
60°
180°
2cos2θ+22sinθ=3 2 \cos ^{2} \theta+2 \sqrt{2} \sin \theta=3 2cos2θ+22sinθ=3
বা, 2(1−sin2θ)+22sinθ=3 2\left(1-\sin ^{2} \theta\right)+2 \sqrt{2} \sin \theta=3 2(1−sin2θ)+22sinθ=3
বা,, (2sinθ−1)2=0 (\sqrt{2} \sin \theta-1)^{2}=0 (2sinθ−1)2=0
বা, 2sinθ−1=0 \sqrt{2} \sin \theta-1=0 2sinθ−1=0
বা, sinθ=12=sin45∘∴θ=45∘ \sin \theta=\frac{1}{\sqrt{2}}=\sin 45^{\circ} \therefore \theta=45^{\circ} sinθ=21=sin45∘∴θ=45∘
f(x)=sinx \mathrm{f}(x)=\sin x f(x)=sinx এবং g(x)=cosx g(x)=\cos x g(x)=cosx.
h(x)=sin−1x \mathrm{h}(\mathrm{x})=\sin ^{-1} \mathrm{x} h(x)=sin−1x এবং p(x)=cosx \mathrm{p}(\mathrm{x})=\cos \mathrm{x} p(x)=cosx
দৃশ্যকল্প-২: 4cosxcos2xcos3x=1 4 \cos x \cos 2 x \cos 3 x=1 4cosxcos2xcos3x=1
g(x)=psin−1x;h(x)=cosx g(x)=p \sin ^{-1} x ; h(x)=\cos x g(x)=psin−1x;h(x)=cosx
