第13章 13.18

(1)

3cos⁡x−sin⁡x=12(32cos⁡x−12sin⁡x)=12sin⁡(π3−x)=1sin⁡(π3−x)=12\sqrt{3}\cos x - \sin x = 1 \qquad 2\left(\frac{\sqrt{3}}{2}\cos x - \frac{1}{2}\sin x\right) = 1 \qquad 2\sin\left(\frac{\pi}{3} - x\right) = 1 \qquad \sin\left(\frac{\pi}{3} - x\right) = \frac{1}{2}

これから

x=π6+2nπ,−π2+2nπx = \frac{\pi}{6} + 2n\pi, -\frac{\pi}{2} + 2n\pi

(2)

sin⁡2x=sin⁡x2sin⁡x⋅cos⁡x=sin⁡xsin⁡x(2cos⁡x−1)=0sin⁡x=0cos⁡x=12\sin 2x = \sin x \qquad 2\sin x \cdot \cos x = \sin x \qquad \sin x(2\cos x - 1) = 0 \qquad \sin x = 0 \qquad \cos x = \frac{1}{2}

これから

x=nπ,±π3+2nπx = n\pi, \pm\frac{\pi}{3} + 2n\pi

(3)

cos⁡2x=sin⁡x+cos⁡xcos⁡2x−sin⁡2x=(cos⁡x−sin⁡x)(cos⁡x+sin⁡x)=sin⁡x+cos⁡x\cos 2x = \sin x + \cos x \qquad \cos^2 x - \sin^2 x = (\cos x - \sin x)(\cos x + \sin x) = \sin x + \cos x

cos⁡x−sin⁡x=1cos⁡x+sin⁡x=0\cos x - \sin x = 1 \qquad \cos x + \sin x = 0

これから

x=−π4+nπ,2nπ,−π2+2nπx = -\frac{\pi}{4} + n\pi, 2n\pi, -\frac{\pi}{2} + 2n\pi

(4)

cos⁡x+cos⁡2x+cos⁡3x=02cos⁡2x⋅cos⁡x+cos⁡2x=0cos⁡2x(2cos⁡x+1)=0\cos x + \cos 2x + \cos 3x = 0 \qquad 2\cos 2x \cdot \cos x + \cos 2x = 0 \qquad \cos 2x(2\cos x + 1) = 0

cos⁡2x=0cos⁡x=−12\cos 2x = 0 \qquad \cos x = -\frac{1}{2}

これから

x=±π4+2nπ,±34π+2nπ,±23π+2nπx = \pm\frac{\pi}{4} + 2n\pi, \pm\frac{3}{4}\pi + 2n\pi, \pm\frac{2}{3}\pi + 2n\pi

解説: r31bn1z