# Solve this trigonometric equation \sin 2x- \sqrt3 \cos 2x=2 I tried

Solve this trigonometric equation $\mathrm{sin}2x-\sqrt{3}\mathrm{cos}2x=2$
I tried dividing both sides with $\mathrm{cos}2x$ but then I win $\frac{2}{\mathrm{cos}2x}$
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sukljama2
$\frac{12}{\mathrm{sin}2}x-\frac{\sqrt{3}}{2}\mathrm{cos}2x=1$
$\mathrm{sin}2x\mathrm{cos}\frac{\pi }{3}-\mathrm{cos}2x\mathrm{sin}\frac{\pi }{3}=1$
$\mathrm{sin}\left(2x-\frac{\pi }{3}\right)=1$
I hope you can solve further.
###### Not exactly what you’re looking for?
abonirali59
MySolution:: Given $\mathrm{sin}2x-\sqrt{3}\mathrm{cos}2x=2$
We can write it as
$\mathrm{sin}2x\cdot \frac{12}{-}\mathrm{cos}2x\cdot \frac{\sqrt{3}}{2}=1⇒\mathrm{sin}\left(2x-\frac{\pi }{3}\right)=1=\mathrm{sin}\frac{\pi }{2}$
Above we have used the formula
$\mathrm{sin}\alpha \cdot \mathrm{cos}\beta -\mathrm{cos}\alpha \cdot \mathrm{sin}\beta =\mathrm{sin}\left(\alpha -\beta \right)$
So
$2x-\frac{\pi }{3}=n\pi +{\left(-1\right)}^{n}\cdot \frac{\pi }{2}⇒x=\frac{n\pi }{2}+{\left(-1\right)}^{n}\cdot \frac{\pi }{4}+\frac{\pi }{6}$
Where $n\in \mathbb{Z}$
###### Not exactly what you’re looking for?
nick1337

Divide by 2 to get
$1=\frac{1}{2}\mathrm{sin}\left(2x\right)-\frac{\sqrt{3}}{2}\mathrm{cos}\left(2x\right)$
then find an angle y for which $\mathrm{cos}y=\frac{1}{2}$ and $\mathrm{sin}y=\frac{\sqrt{3}}{2}$ and remember that
$\mathrm{sin}\left(a-b\right)=\mathrm{sin}a\mathrm{cos}b-\mathrm{cos}a\mathrm{sin}b$