Question

Solve $\sin x+\sqrt{3}\cos x=\sqrt{2}$

Answer 1

Solution:-

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

$\sqrt{3}\cos x=\sqrt{2}-\sin x$

Squaring both sides, we have

${\left(\sqrt{3}\cos x\right)}^2={(\sqrt{2}-\sin x)}^2$

$3{\cos }^{2}x=2+{\sin }^{2}x-2\sqrt{2}\sin x$

$\Rightarrow 3(1-{\sin }^{2}x)=2+{\sin }^{2}x-2\sqrt{2}\sin x(\because {\sin }^{2}x+{\cos }^{2}x=1)$

$3-3{\sin }^{2}x=2+{\sin }^{2}x-2\sqrt{2}\sin x$

$\Rightarrow 4{\sin }^{2}x-2\sqrt{2}\sin x-1=0$

The above equation is in the quadratic form, i.e. $a{x}^2+bx+c$.

Now, by using quadratic formula, i.e., $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, we have

$\sin x=\frac{-(-2\sqrt{2})\pm \sqrt{{(-2\sqrt{2})}^2-(4\times 4\times (-1))}}{2\times 4}$

$\Rightarrow \sin x=\frac{2\sqrt{2}\pm \sqrt{8+16}}{8}$

$\Rightarrow \sin x=\frac{\sqrt{2}\pm \sqrt{6}}{4}$

$\therefore x={\sin }^{-1}\left(\frac{\sqrt{2}+\sqrt{6}}{4}\right)+2n\pi \text{Or}x={\sin }^{-1}\left(\frac{\sqrt{2}-\sqrt{6}}{4}\right)+2n\pi \text{For all n}\in \text{integer}$