How to Simplify $-\frac14\sin\frac x4 + \frac{\sqrt{3}}4\cos\frac x4 = 0$? A problem requires finding the value of x, such that 
$$-\frac14\sin\frac x4 + \frac{\sqrt{3}}4\cos\frac x4 = 0$$
The site reduces the problem to $\tan(\frac x4) = \sqrt{3}$.
How one should do this? 
 A: Hint:
It is the expansion of $\;\displaystyle\frac12\cos\Bigl(\frac x4 +\frac\pi6\Bigr)$.
A: Another approach is to use $\sin(a-b)=\sin{a}\cos{b}-\cos{a}\sin{b}$ and we get our equation after having factored ${1\over 2}$ by setting $b={\pi\over 3}$ and $a={x\over 4}$ so we have to solve
$${1\over 2}\sin\left({x \over 4}-{\pi\over 3}\right)=0$$
And this leads to the result as expected
A: Method 1
$$\begin{align}
-\frac14\sin\frac x4 + \frac{\sqrt{3}}4\cos\frac x4 &= 0 \\
-\frac14\sin\frac x4 &= - \frac{\sqrt{3}}4\cos\frac x4 & \text{Take cos to the other side}\\
\sin\frac x4 &= \sqrt{3} \cos \frac x4 & \text{Multiply by -4} \\
\tan \frac x4 &= \sqrt{3} & \text{Devide by cos}
\end{align}$$
But in this method you should take care that $\cos \frac x4 \ne 0$ as you devided by it at the last step. But happily $\cos \frac x4 = 0$ will never happen! (Why?)
Method 2
$$\begin{align}
-\frac14\sin\frac x4 + \frac{\sqrt{3}}4\cos\frac x4 &= 0 \\
-\frac12 \left(\frac 12 \sin\frac x4 - \frac{\sqrt{3}}2\cos\frac x4 \right) &= 0  & \text{Factor $-\frac {1} {2}$}\\
-\frac12 \left(\cos \frac \pi 3 \sin\frac x4 - \sin \frac \pi 3 \cos\frac x4 \right) &= 0 & \text{Note that $\cos \frac \pi 3 = \frac 1 2$ and $\sin \frac \pi 3 = \frac {\sqrt{3}} {2}$} \\
-\frac{1}{2} \sin \left( \frac x 4 - \frac \pi 3 \right) &= 0 & \text{Use summation formula for sin} \\
\sin \left( \frac x 4 - \frac \pi 3 \right) &=0
\end{align}$$
You can take a look at this post for a more general case of method $2$.
Solution of the Equation
Using either method $1$ or method $2$ one will conclude that
$$x=\frac{4\pi}{3}+4n\pi, \qquad \qquad n=0,\pm1,\pm2,...$$
A: Simplify:
$$
\begin{aligned}
-\frac14\sin\frac x4 + \frac{\sqrt3}4\cos\frac x4 &=\frac12\left(-\frac12\sin\frac x4 + \frac{\sqrt{3}}2\cos\frac x4\right)\\
&=\frac12\left(-\sin\frac\pi6\sin\frac x4+\cos\frac\pi6\cos\frac x4\right)\\
&=\frac12\cos(\frac\pi6+\frac x4)
\end{aligned}
$$
A: $$-\frac { 1 }{ 4 } \sin  \frac { x }{ 4 } +\frac { \sqrt { 3 }  }{ 4 } \cos  \frac { x }{ 4 } =0\\ -\frac { 1 }{ 2 } \left( \frac { 1 }{ 2 } \sin  \frac { x }{ 4 } -\frac { \sqrt { 3 }  }{ 2 } \cos  \frac { x }{ 4 }  \right) =0\\ \frac { 1 }{ 2 } \sin  \frac { x }{ 4 } -\frac { \sqrt { 3 }  }{ 2 } \cos  \frac { x }{ 4 } =0\\ \cos { \frac { \pi  }{ 3 } \sin { \frac { x }{ 4 } -\sin { \frac { \pi  }{ 3 }  } \cos { \frac { x }{ 4 } =0 }  }  } \\ \sin { \left( \frac { x }{ 4 } -\frac { \pi  }{ 3 }  \right) =0 } \\ x=\frac { 4\pi  }{ 3 } +4k\pi ,k\epsilon  \mathbb{Z}  $$
A: $$-\frac{1}{4}\sin\left(\frac{x}{4}\right)+\frac{\sqrt{3}}{4}\cos\left(\frac{x}{4}\right)=0\Longleftrightarrow$$
$$\frac{1}{4}\left(\sqrt{3}\cos\left(\frac{x}{4}\right)-\sin\left(\frac{x}{4}\right)\right)=0\Longleftrightarrow$$
$$\sqrt{3}\cos\left(\frac{x}{4}\right)-\sin\left(\frac{x}{4}\right)=0\Longleftrightarrow$$
$$\sqrt{3}\cos\left(\frac{x}{4}\right)=\sin\left(\frac{x}{4}\right)\Longleftrightarrow$$
$$\sqrt{3}\cot\left(\frac{x}{4}\right)=1\Longleftrightarrow$$
$$\cot\left(\frac{x}{4}\right)=\frac{1}{\sqrt{3}}\Longleftrightarrow$$
$$\frac{x}{4}=\pi n+\text{arccot}\left(\frac{1}{\sqrt{3}}\right)\Longleftrightarrow$$
$$x=4\pi n+4\text{arccot}\left(\frac{1}{\sqrt{3}}\right)\Longleftrightarrow$$
$$x=4\pi n+\frac{4\pi}{3}\space\space\space\space\space\space\space\space\text{with}\space\space n\in\mathbb{Z}$$
