1

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?

Hosein Rahnama
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Haim
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6 Answers6

2

Hint:

It is the expansion of $\;\displaystyle\frac12\cos\Bigl(\frac x4 +\frac\pi6\Bigr)$.

Bernard
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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

marwalix
  • 16,773
2

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

$$x=\frac{4\pi}{3}+4n\pi, \qquad \qquad n=0,\pm1,\pm2,...$$

Community
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Hosein Rahnama
  • 14,882
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    In method 1, you can't have both sine and cosine equal to $0$. In method 2, you should add some pairs of parentheses at the end. – Bernard Dec 26 '15 at 20:43
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Simplify: $$ \begin{aligned} -\frac14\sin\frac x4 + \frac{\sqrt3}4\cos\frac x4 &=\frac12\left(-\frac12\sin\frac x4 + \frac{\sqrt{3}}2\cos\frac x4\right)\\ &=\frac12\left(-\sin\frac\pi6\sin\frac x4+\cos\frac\pi6\cos\frac x4\right)\\ &=\frac12\cos(\frac\pi6+\frac x4) \end{aligned} $$

Jyrki Lahtonen
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$$-\frac { 1 }{ 4 } \sin \frac { x }{ 4 } +\frac { \sqrt { 3 } }{ 4 } \cos \frac { x }{ 4 } =0\\ -\frac { 1 }{ 2 } \left( \frac { 1 }{ 2 } \sin \frac { x }{ 4 } -\frac { \sqrt { 3 } }{ 2 } \cos \frac { x }{ 4 } \right) =0\\ \frac { 1 }{ 2 } \sin \frac { x }{ 4 } -\frac { \sqrt { 3 } }{ 2 } \cos \frac { x }{ 4 } =0\\ \cos { \frac { \pi }{ 3 } \sin { \frac { x }{ 4 } -\sin { \frac { \pi }{ 3 } } \cos { \frac { x }{ 4 } =0 } } } \\ \sin { \left( \frac { x }{ 4 } -\frac { \pi }{ 3 } \right) =0 } \\ x=\frac { 4\pi }{ 3 } +4k\pi ,k\epsilon \mathbb{Z} $$

haqnatural
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$$-\frac{1}{4}\sin\left(\frac{x}{4}\right)+\frac{\sqrt{3}}{4}\cos\left(\frac{x}{4}\right)=0\Longleftrightarrow$$ $$\frac{1}{4}\left(\sqrt{3}\cos\left(\frac{x}{4}\right)-\sin\left(\frac{x}{4}\right)\right)=0\Longleftrightarrow$$ $$\sqrt{3}\cos\left(\frac{x}{4}\right)-\sin\left(\frac{x}{4}\right)=0\Longleftrightarrow$$ $$\sqrt{3}\cos\left(\frac{x}{4}\right)=\sin\left(\frac{x}{4}\right)\Longleftrightarrow$$ $$\sqrt{3}\cot\left(\frac{x}{4}\right)=1\Longleftrightarrow$$ $$\cot\left(\frac{x}{4}\right)=\frac{1}{\sqrt{3}}\Longleftrightarrow$$ $$\frac{x}{4}=\pi n+\text{arccot}\left(\frac{1}{\sqrt{3}}\right)\Longleftrightarrow$$ $$x=4\pi n+4\text{arccot}\left(\frac{1}{\sqrt{3}}\right)\Longleftrightarrow$$ $$x=4\pi n+\frac{4\pi}{3}\space\space\space\space\space\space\space\space\text{with}\space\space n\in\mathbb{Z}$$

Jan Eerland
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