Laney Spears

2022-01-29

How does one solve $\mathrm{sin}x-\sqrt{3}\mathrm{cos}x=1$
I thought this one up, but I am not sure how to solve it. Here is my attempt:
$\mathrm{sin}x-\sqrt{3}\mathrm{cos}x=1$
${\left(\mathrm{sin}x-\sqrt{3}\mathrm{cos}x\right)}^{2}=1$
${\mathrm{sin}}^{2}x-2\sqrt{3}\mathrm{sin}x\mathrm{cos}x+3{\mathrm{cos}}^{2}x=1$
$1-2\sqrt{3}\mathrm{sin}x\mathrm{cos}x+2{\mathrm{cos}}^{2}x=1$
$2{\mathrm{cos}}^{2}x-2\sqrt{3}\mathrm{sin}x\mathrm{cos}x=0$
$2\mathrm{cos}x\left(\mathrm{cos}x-\sqrt{3}\mathrm{sin}x\right)=0$
$2\mathrm{cos}x=0⇒x\in \left\{\frac{\pi }{2}\left(2n-1\right):n\in \mathbb{Z}\right\}$
But how do I solve
$\mathrm{cos}x-\sqrt{3}\mathrm{sin}x=0$

chukizosv

Expert

Hint: at the very beginning divide both sides by 2 and use the formula for the sin of difference of 2 arguments

vasselefa

Expert

Hint :
$\mathrm{cos}x-\sqrt{3}\mathrm{sin}x=0⇔\frac{\mathrm{sin}x}{\mathrm{cos}x}=\frac{\sqrt{3}}{3}⇔\mathrm{tan}x=\frac{\sqrt{3}}{3}$
Note : You can divide by $\mathrm{cos}x$ since if the case was $\mathrm{cos}x=0$ it would be $\mathrm{sin}x=±1$ and thus the equation would yield $±\sqrt{3}±0$, thus no problems in the final solution, as the cos zeros are no part of it.

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