Question

Sin3x*cos4x+cos3x*sin4x=√3\2​

Answer 1

$\sin 3x\cos 4x+\cos 3x\sin4x=\dfrac{\sqrt{3}}{2}\\ \\ \sin(3x+4x)=\dfrac{\sqrt{3}}{2}\\ \\ \sin7x=\dfrac{\sqrt{3}}{2}\\ \\ 7x=(-1)^k\cdot \dfrac{\pi}{3}+\pi k,k \in \mathbb{Z}\\ \\ x=(-1)^k\cdot \dfrac{\pi}{21}+\dfrac{\pi k}{7},k \in \mathbb{Z}$