sin(3x)=cos(5x)

sin(3x)=cos(5x)

how to solve this

\(\small{ \begin{array}{rcl} &\text{Formula } \\ &\boxed{~ \begin{array}{rcl} \cos{(A)}-\cos{(B)} &=& -2 \cdot \sin{(\frac{A+B}{2})} \cdot \sin{(\frac{A-B}{2})}\\ \end{array}\\ ~}\\\\ \end{array}\\ \begin{array}{lrcll} & \sin{(3x)}&=& \cos{(5x)} \\ & \sin{(3x)}- \cos{(5x)}&=& 0 \qquad | \qquad \sin{(3x)}=\cos{( \frac{\pi}{2}-3x )} \\ & \cos{( \underbrace{ \frac{\pi}{2}-3x }_{=A} )}- \cos{( \underbrace{ 5x }_{=B} )}&=& 0 \\ & \cos{(\frac{\pi}{2}-3x )}- \cos{( 5x )} &=& -2\cdot \sin{(\frac{\frac{\pi}{2}-3x+5x }{2})} \cdot \sin{(\frac{\frac{\pi}{2}-3x-(5x) }{2})} \qquad \text{Formula}\\ & &=& -2\cdot \sin{(\frac{\frac{\pi}{2}+2x }{2})} \cdot \sin{(\frac{\frac{\pi}{2}-8x }{2})}\\\\ & \mathbf{ \sin{(3x)}- \cos{(5x)} } & \mathbf{=} & \mathbf{ -2\cdot \sin{ ( \frac{\pi}{4}+x) } \cdot \sin{( \frac{\pi}{4}-4x)} = 0 }\\\\ & -2\cdot \sin{ ( \frac{\pi}{4}+x) } \cdot \sin{( \frac{\pi}{4}-4x)} &=& 0 \qquad | \qquad : -2\\ & \underbrace{ \sin{ ( \frac{\pi}{4}+x) } }_{=0} \cdot \underbrace{ \sin{( \frac{\pi}{4}-4x)} }_{=0}&=& 0\\\\ \hline 1. \text{ Solution} & \sin{ ( \frac{\pi}{4}+x ) } &=& 0 \\ & \frac{\pi}{4}+x &=& \arcsin{ ( 0 ) } \pm 2k\cdot \pi \\ & \frac{\pi}{4}+x &=& \pm 2k\cdot \pi \\ & \mathbf{x} &\mathbf{=}&\mathbf{ - \frac{\pi}{4} \pm 2k\cdot \pi } \qquad k \in Z\\ \hline 2. \text{ Solution} & \sin{ ( \pi - ( \frac{ \pi}{4}+x ) ) } &=& 0 \\ & \sin{ ( \frac{3 \pi}{4}-x ) } &=& 0 \\ & \frac{3 \pi}{4}-x &=& \arcsin{ ( 0 ) } \pm 2k\cdot \pi \\ & \frac{3 \pi}{4}-x &=& \pm 2k\cdot \pi \\ & \mathbf{x} &\mathbf{=}& \mathbf{\frac{3 \pi}{4} \pm 2k\cdot \pi } \qquad k \in Z\\ \hline 3. \text{ Solution} & \sin{ ( \frac{\pi}{4}-4x ) } &=& 0\\ & \frac{\pi}{4}-4x &=& \arcsin{ ( 0 ) } \pm 2k\cdot \pi \\ & \frac{\pi}{4}-4x &=& \pm 2k\cdot \pi \\ & 4x &=& \frac{\pi}{4} \pm 2k\cdot \pi \\ & \mathbf{x} &\mathbf{=}& \mathbf{\frac{\pi}{16} \pm k\cdot \frac{\pi}{2} } \qquad k \in Z\\ \hline 4. \text{ Solution} & \sin{ ( \pi - ( \frac{ \pi}{4}-4x ) ) } &=& 0\\ & \sin{ ( \frac{3\pi}{4}+4x ) } &=& 0\\ & \frac{3\pi}{4}+4x &=& \arcsin{ ( 0 ) } \pm 2k\cdot \pi \\ & \frac{3\pi}{4}+4x &=& \pm 2k\cdot \pi \\ & 4x &=& -\frac{3\pi}{4} \pm 2k\cdot \pi\\ & \mathbf{x} &\mathbf{=}& \mathbf{ -\frac{3\pi}{16} \pm k\cdot \frac{\pi}{2} } \qquad k \in Z\\ \end{array} }\)