Question

Find $\theta$, if $\sin \theta +\cos \theta =\frac{\sqrt{3}+1}{2}$ and $\sqrt{3}\sin \theta +4\cos \theta =\frac{7}{2}$

Answer 1

$\sin \theta +\cos \theta =\frac{\sqrt{3}+1}{2}$

$\sin \theta =\frac{\sqrt{3}+1}{2}-\cos \theta$

$\sqrt{3}\sin \theta +4\cos \theta =\frac{7}{2}$

$\sqrt{3}(\frac{\sqrt{3}+1}{2}-\cos \theta )+4\cos \theta =\frac{7}{2}$

$3+\sqrt{3}-2\sqrt{3}\cos \theta +8\cos \theta =7$

$(8-2\sqrt{3})\cos \theta =4-\sqrt{3}$

$\cos \theta =\frac{1}{2}$

$\theta =\frac{\pi }{3}$