Find the** general **solution of the equation:

Sin Ø - Sin 2Ø = Sin 4Ø - Sin 3Ø

Answer :2np,(2n - 1)p/2,(2n + 1)p/5 (p=pye)

OldTimer Dec 8, 2018

#4**+1 **

Hi Alan...great job thanks!

Prior to posting question and after long hours of problem solving I did obtain 36 deg and 108 degrees. My main issue was applying the standard formulas for the general angle...in this case cosine. Therefore I am now trying to see how I can reconcile those answers....hopefully I can solve !

All the best for Xmas of course and New Year....thanks again!

OldTimer Dec 9, 2018

#5

#6**0 **

Hi Alan thanks for your follow up......If I may bother you for a bit more clarification

>>The formulas i am following are as below:

Therefore, I have understood solutions obtained as far as angles go...

**but**

the logic of how you derived extensions (as **highlighted** below in your answer) escapes me.

Thanks for your patience!

OldTimer Dec 10, 2018

#8**+9 **

**General Solution - Trigonometry**

**Sin Ø - Sin 2Ø = Sin 4Ø - Sin 3Ø**

\(\begin{array}{ rcl } \sin(\phi) - \sin(2\phi) &=& \sin(4\phi) - \sin(3\phi) \\ &\text{or} & \\ \sin(3\phi)+\sin(\phi) &=& \sin(4\phi) + \sin(2\phi) \\ \end{array} \)

**Formula:**

\(\begin{array}{|rcll|} \hline \underbrace{\sin(3\phi)+\sin(\phi)}_{=2\sin(2\phi)\cos(\phi)} &=& \underbrace{\sin(4\phi) + \sin(2\phi)}_{=2\sin(3\phi)\cos(\phi)} \\\\ 2\sin(2\phi)\cos(\phi) &=& 2\sin(3\phi)\cos(\phi) \\\\ \sin(2\phi)\cos(\phi) &=& \sin(3\phi)\cos(\phi) \\\\ \sin(3\phi)\cos(\phi) -\sin(2\phi)\cos(\phi) &=& 0 \\\\ \cos(\phi)\Big(\sin(3\phi)-\sin(2\phi) \Big) &=& 0 \\ \hline \end{array} \)

**Formula:**

\(\begin{array}{|rcll|} \hline \cos(\phi)\Big(\underbrace{\sin(3\phi)-\sin(2\phi)}_{=2\cos(\frac52\phi)\sin(\frac12\phi)} \Big) &=& 0 \\\\ \cos(\phi)2\cos(\frac52\phi)\sin(\frac12\phi) &=& 0 \quad | \quad :2 \\\\ \large{\mathbf{\cos(\phi)\cos(\frac52\phi)\sin(\frac12\phi)} }& \large{\mathbf{=}} & \large{\mathbf{0}} \\ \hline \end{array}\)

**General Solution:**

\(\begin{array}{|lrcll|} \hline 1. & \cos(\phi) &=& 0 \\ & \phi &=& 2n\pi\pm\arccos(0) \\ & \phi &=& 2n\pi\pm \frac{\pi}{2} \quad \Rightarrow \quad \mathbf{\phi = (2n-1)\frac{\pi}{2}} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline 2. & \cos(\frac52\phi) &=& 0 \\ & \frac52\phi &=& 2n\pi\pm\arccos(0) \\ & \frac52\phi &=& 2n\pi\pm \frac{\pi}{2}\quad | \quad \cdot \frac25 \\ & \phi &=& \frac45 n\pi\pm \frac{\pi}{5} \quad \Rightarrow \quad \mathbf{\phi = (2n-1)\frac{\pi}{5}} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline 3. & \sin(\frac12\phi) &=& 0 \\ & \frac12\phi &=& n\pi+(-1)^n \arcsin(0) \\ & \frac12\phi &=& n\pi+(-1)^n \pi \quad | \quad \cdot 2 \\ & \phi &=& 2n\pi+(-1)^n 2\pi \quad \Rightarrow \quad \mathbf{\phi = 2n\pi} \\ \hline \end{array} \)

heureka Dec 12, 2018