+8 Solve Sin(3x)+sin(4x)=0
+118608 Not so easy.
Here is the graph of y=sin(3x)+sin(4x)
The solutions are the intersections of this graph with y=0 (the x axis)
You can see that there are 9 solutions. (including x=0 and x=2pi)
Trig identity
\(\displaystyle \sin(A)+\sin(B)=2\sin(\frac{A+B}{2})\cos(\frac{A-B}{2})\).
So,
\(\displaystyle \sin(3x)+\sin(4x)=2\sin(\frac{7x}{2})\cos(\frac{x}{2})\),
so
\(\displaystyle \sin(\frac{7x}{2})=0,\text{ or }\cos(\frac{x}{2})=0\)
etc..
+26367 Solve for
\( 0 \text{ to } 2\pi\)
\(\sin(3x)+\sin(4x)=0 \)
\(\begin{array}{|lrcll|} \hline & \sin(3x)+\sin(4x) &=& 0 \quad | \quad -\sin(3x) \\ & \sin(4x) &=& -\sin(3x) \quad | \quad -\sin(3x) = \sin(-3x) \\ & \sin(4x) &=& \sin(-3x) \quad | \quad \arcsin() \\\\ (1) & 4x + 2\pi n &=& -3x + 2\pi m \quad | \quad +3x \\ & 7x &=& 2\pi z \\ & \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{2\pi}{7} z } \qquad z \in \mathbb{Z} \\\\ (2) & \sin(4x) &=& \sin(-3x) \quad | \quad \sin(-3x) = \sin(\pi -(-3x) ) \\ & \sin(4x ) &=& \sin(\pi + 3x ) \quad | \quad \arcsin() \\ & 4x +2\pi n &=& \pi+3x + 2\pi m \quad | \quad -3x \\ & x &=& \pi + 2\pi z \\ & \mathbf{x} &\mathbf{=}& \mathbf{ \pi + 2\pi z } \qquad z \in \mathbb{Z} \\ \hline \end{array}\)
\(\begin{array}{|r|r|r|} \hline & 0\le x\le2\pi & 0\le x\le2\pi \\ z & x=\dfrac{2\pi}{7} z & x = \pi + 2\pi z \\ \hline 0 & 0 & \pi \\ 1 & \dfrac{2\pi}{7} & \\ 2 & \dfrac{4\pi}{7} & \\ 3 & \dfrac{6\pi}{7} & \\ 4 & \dfrac{8\pi}{7} & \\ 5 & \dfrac{10\pi}{7} & \\ 6 & \dfrac{12\pi}{7} & \\ 7 & \dfrac{14\pi}{7}=2\pi & \\ \hline \end{array}\)
\(x = \left\{ 0,\ \dfrac{2\pi}{7},\ \dfrac{4\pi}{7},\ \dfrac{6\pi}{7},\ \pi ,\ \dfrac{8\pi}{7},\ \dfrac{10\pi}{7},\ \dfrac{12\pi}{7},\ 2\pi \right\}\)
