JEE Advanced 2018 — Paper 1 — Question 39

$\sqrt{3} a \cos x+2 b \sin x=c$

$\sqrt{3} a \cos \left(\frac{\pi}{3}-x\right)+2 b \sin \left(\frac{\pi}{3}-x\right)=c$

$\Rightarrow \sqrt{3} \mathrm{a}\left(\cos \mathrm{x} \cdot \frac{1}{2}+\frac{\sin \mathrm{x} \sqrt{3}}{2}\right)+2 \mathrm{~b}\left(\frac{\sqrt{3}}{2} \cos \mathrm{x}-\frac{1}{2} \sin \mathrm{x}\right)=\mathrm{c}$

$\Rightarrow\left(\frac{\sqrt{3}}{2} a+\sqrt{3} b\right) \cos x+\left(\frac{3}{2} a-b\right) \sin x=c$

$\left(\sqrt{3} b-\frac{\sqrt{3}}{2} a\right) \cos x+\left(\frac{3}{2} a-3 b\right) \sin x=0$

$\Rightarrow \frac{\mathrm{b}}{\mathrm{a}}=\frac{1}{2}$