√3.cosX – sinX = √3, How To Solve This?

Hi,

There's a simpler way.

When you don't know where to start, when given $a\cos(x)+b\sin(x)$ , find the modulus, that is $\sqrt{a^2+b^2}$ and factor it out.
You would then have $\sqrt{a^2+b^2} \left(\frac{a}{\sqrt{a^2+b^2}} \cos(x)+\frac{b}{\sqrt{a^2+b^2}} \sin(x)\right)$

Define $c=\frac{a}{\sqrt{a^2+b^2}}$ and $d=\frac{b}{\sqrt{a^2+b^2}}$
they're both in $[-1,1]$ and are such that $c^2+d^2=1$
Thus there exists $\theta \in [0,2\pi[$ such that $c=\cos \theta$ and $d=\sin \theta$

Hence $a\cos x+b\sin x=\cos\theta\cos x+\sin\theta\sin x=\cos(x-\theta)$

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Here, you can find that $\sqrt{a^2+b^2}=2$

So you can divide by 2 on both sides :
$\frac{\sqrt{3}}{2}\cos(x)-\frac 12 \sin(x)=\frac{\sqrt{3}}{2}$

Note that $\cos\left(-\tfrac\pi 6\right)=\tfrac{\sqrt{3}}{2}$ and $\sin\left(-\tfrac\pi 6\right)=-\tfrac 12$

So your equation is now $\cos\left(-\tfrac\pi 6\right)\cos(x)+\sin\left(-\tfrac\pi 6\right)\sin(x)=\tfrac{\sqrt{3}}{2}$

That is $\cos\left(x+\tfrac\pi 6\right)=\tfrac{\sqrt{3}}{2}$

Then remember that $\cos(a)=\cos(b) \Leftrightarrow \forall k\in \mathbb{Z} ~,~ a=\pm b+2k\pi$