Solve this trigonometric equation $\sin2x-\sqrt3\cos2x=2$

Solve equation: $$\sin2x-\sqrt3\cos2x=2$$ I tried dividing both sides with $\cos2x$ but then I win $\frac{2}{\cos2x}$.

• Hint: Use $\sin(a-b) = \sin a \cos b - \sin b \cos b$ to rewrite the left hand side as $A\sin(2x - c)$ for suitable constants $A$ and $c$. Aug 9, 2015 at 12:32

$$\frac{1}{2}\sin 2x-\frac{\sqrt3}{2}\cos 2x=1$$

$$\sin 2x \cos \frac{\pi}{3}-\cos 2x\sin \frac{\pi}{3}=1$$

$$\sin(2x-\frac{\pi}{3})=1$$

I hope you can solve further.

• Yes, I can, thank you! Aug 9, 2015 at 12:34

$\bf{My\; Solution::}$ Given $\displaystyle \sin 2x -\sqrt{3}\cos 2x = 2$

We can write it as $$\displaystyle \sin 2x \cdot \frac{1}{2}-\cos 2x\cdot \frac{\sqrt{3}}{2} = 1\Rightarrow \sin \left(2x-\frac{\pi}{3}\right) =1=\sin \frac{\pi}{2}$$

Above we have used the formula $$\sin \alpha\cdot \cos \beta-\cos \alpha\cdot \sin \beta=\sin (\alpha-\beta).$$

So $$\displaystyle 2x-\frac{\pi}{3} = n\pi+(-1)^n\cdot \frac{\pi}{2}\Rightarrow x=\frac{n\pi}{2}+(-1)^n\cdot \frac{\pi}{4}+\frac{\pi}{6}\;,$$ Where $n\in \mathbb{Z}$

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

Can you finish this?

$$\sin(2x)-\sqrt3\cos(2x)=2\Longleftrightarrow$$ $$-2\cos\left(2x+\frac{\pi}{6}\right)=2\Longleftrightarrow$$ $$\cos\left(2x+\frac{\pi}{6}\right)=-1\Longleftrightarrow$$ $$2x+\frac{\pi}{6}=\pi\Longleftrightarrow$$ $$2x=\pi - \frac{\pi}{6}\Longleftrightarrow$$ $$2x=\frac{5\pi}{6}\Longleftrightarrow$$ $$x=\frac{\frac{5\pi}{6}}{2}\Longleftrightarrow$$ $$x=\frac{5\pi}{12}$$

So looking to the integer solutions we found:

$$x=\frac{12\pi n +5\pi}{12}, n \in \mathbb{Z}$$

$$\sin2x-\sqrt3\cos2x=2$$

\begin{align}=-2\bigg(\frac12 \sqrt3 \cos(2x)-\frac12 \sin(2x)\bigg)\\=\bigg(\cos \big(\frac{\pi}{6}\big) \cos(2x)-\sin\big(\frac{\pi}{6}\big) \sin(2x)\bigg)\\=-2\cos\bigg(\frac{\pi}{6}+2x\bigg)=2\\=\cos\bigg(\frac{\pi}{6}+2x\bigg)=-1\\ \boxed{x=\frac{5 \pi}{12}+\pi n \;\;\;\;\;\;\;\;\;\; n\in \mathbb{Z}}\end{align}

Notice, we have $$\sin 2x-\sqrt 3\cos 2x=2$$ $$\frac{1}{2}\sin 2x-\frac{\sqrt 3}{2}\cos 2x=1$$ $$\sin 2x\cos\frac{\pi}{3}-\cos 2x\sin \frac{\pi}{3}=1$$ $$\sin\left(2x-\frac{\pi}{3}\right)=1=\sin \frac{\pi}{2}$$ Now, writing the general solution for $x$, we get $$2x-\frac{\pi}{3}=2n\pi+\frac{\pi}{2}$$ $$2x=2n\pi+\frac{\pi}{2}+\frac{\pi}{3}=2n\pi+\frac{5\pi}{6}$$

$$\bbox[5px, border:2px solid #C0A000]{\color{red}{x=n\pi+\frac{5\pi}{12}}}$$ Where, $n$ is any integer

$sin(2x)−\sqrt 3 cos(2x)=R sin(2x-\alpha)$; $R=\sqrt{1+3}=2$, $tan \alpha=\frac{1}{\sqrt 3}$, $\alpha=\frac{\pi}{6}$.

$R sin(2x-\alpha)=2 sin(2x-\frac{\pi}{6})=2$

$sin(2x-\frac{\pi}{6})=1$

$(2x-\frac{\pi}{6})=\frac{\pi}{2}$

$2x=\frac{2\pi}{3}$

Hence, $x=\frac{\pi}{3}$,... (depending on limits you have for $x$)

• You can use $\sin 2x$ to typeset $\sin 2x$ and $\sqrt{1+3}$ to typeset $\sqrt{1+3}$. Jan 8, 2016 at 9:19