sin3x+sin(x+π2)=2sin(2x+π4)cos(x−π4)=2sin(2x+π4)sin(x+π4)=3sin(x+π4)\sin 3x + \sin\left(x+\dfrac{\pi}{2}\right) = 2\sin\left(2x+\dfrac{\pi}{4}\right)\cos\left(x-\dfrac{\pi}{4}\right) = 2\sin\left(2x+\dfrac{\pi}{4}\right)\sin\left(x+\dfrac{\pi}{4}\right) = \sqrt{3}\sin\left(x+\dfrac{\pi}{4}\right)sin3x+sin(x+2π)=2sin(2x+4π)cos(x−4π)=2sin(2x+4π)sin(x+4π)=3sin(x+4π) sin(x+π4)=0\sin\left(x+\dfrac{\pi}{4}\right) = 0sin(x+4π)=0 これから −π4+nπ-\dfrac{\pi}{4}+n\pi−4π+nπ sin(x+π4)≠0\sin\left(x+\dfrac{\pi}{4}\right) \neq 0sin(x+4π)=0 のとき sin(2x+π4)=32\sin\left(2x+\dfrac{\pi}{4}\right) = \dfrac{\sqrt{3}}{2}sin(2x+4π)=23 これから π24+nπ\dfrac{\pi}{24}+n\pi24π+nπ、524π+nπ\dfrac{5}{24}\pi+n\pi245π+nπ