If $sin \alpha + sin \beta = \dfrac{1}{2}$ and $cos \alpha + cos \beta = \dfrac{\sqrt{3}}{2}$ then $3 \beta + \alpha =$

এখানে ,

$\sin \alpha+\sin \beta=1 / 2$

$\begin{array}{l} \Rightarrow 2 \sin \frac{\alpha+\beta}{2} \cdot \cos \frac{\alpha-\beta}{2}=1 / 2 \\ \cos \alpha+\cos \beta=\sqrt{3} / 2 \\ \Rightarrow 2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}=\sqrt{3} / 2 \\ \\ \tan \frac{\alpha+\beta}{2}=\frac{1 / 2}{\sqrt{3} / 2} \\ \Rightarrow 1 / \sqrt{3} \\ \Rightarrow \frac{\alpha+\beta}{2}=\frac{\pi}{6} \\ \Rightarrow \alpha+\beta=\pi / 3 \end{array}$

$\begin{array}{c} \sin ^{2} \alpha+2 \sin \alpha \sin \beta+\sin ^{2} \beta \\ +\cos ^{2} \alpha+2 \cos \alpha \cdot \cos \beta+\cos ^{2} \beta \\ =1 / 4+3 / 4 \\ 1+1+28(\cos \alpha \cos \beta \end{array}$

$\Rightarrow 1+1+28(\cos \alpha \cos \beta$

$+\sin \alpha \cdot \sin \beta)=1$

$\begin{array}{l} \Rightarrow \quad \ ২cos (\beta-\alpha)=1-2 \\ \Rightarrow \cos (\beta-\alpha)=-1 / 2 \\ \Rightarrow \quad \beta-\alpha=2 \pi / 3 \end{array}$

$\alpha=-\pi / 6 \quad \beta=\pi / 2$

প্রদত্ত রশ্মি,

$\begin{aligned} & 3 \alpha+\beta \\ \Rightarrow & 3 \times(-\pi / 6)+\pi / 2 \\ \Rightarrow & -\pi / 2+\pi / 2 \\ = & 0 \\ & \end{aligned}$