第13章 13.10

(1)

sin⁡x−1=0sin⁡x=1x=π2,52π,⋯x=π2+2nπ\sin x - 1 = 0 \qquad \sin x = 1 \qquad x = \frac{\pi}{2}, \frac{5}{2}\pi, \cdots \qquad x = \frac{\pi}{2} + 2n\pi

(2)

2sin⁡x+1=0sin⁡x=−12x=76π+2nπ,−π6+2nπ2\sin x + 1 = 0 \qquad \sin x = -\frac{1}{2} \qquad x = \frac{7}{6}\pi + 2n\pi, -\frac{\pi}{6} + 2n\pi

(3)

tan⁡x+1=0tan⁡x=−1x=3π4+nπ\tan x + 1 = 0 \qquad \tan x = -1 \qquad x = \frac{3\pi}{4} + n\pi

(4)

cos⁡x−2cos⁡2x=0cos⁡x(1−2cos⁡x)=0cos⁡x=0,12x=π2+nπ,±π3+2nπ\cos x - 2\cos^2 x = 0 \qquad \cos x(1 - 2\cos x) = 0 \qquad \cos x = 0, \frac{1}{2} \qquad x = \frac{\pi}{2} + n\pi, \pm\frac{\pi}{3} + 2n\pi

(5)

2sin⁡2x+sin⁡x−1=0(2sin⁡x−1)(sin⁡x+1)=0sin⁡x=12,−12\sin^2 x + \sin x - 1 = 0 \qquad (2\sin x - 1)(\sin x + 1) = 0 \qquad \sin x = \frac{1}{2}, -1

x=π6+2nπ,56π+2nπ,32π+2nπx = \frac{\pi}{6} + 2n\pi, \frac{5}{6}\pi + 2n\pi, \frac{3}{2}\pi + 2n\pi

(6)

cos⁡2x−3cos⁡x+2=0(cos⁡x−2)(cos⁡x−1)=0cos⁡x=1x=2nπ\cos^2 x - 3\cos x + 2 = 0 \qquad (\cos x - 2)(\cos x - 1) = 0 \qquad \cos x = 1 \qquad x = 2n\pi

(7)

sin⁡2x+sin⁡x=02sin⁡xcos⁡x+sin⁡x=0sin⁡x(2cos⁡x+1)=0sin⁡x=0cos⁡x=−12\sin 2x + \sin x = 0 \qquad 2\sin x \cos x + \sin x = 0 \qquad \sin x(2\cos x + 1) = 0 \qquad \sin x = 0 \qquad \cos x = -\frac{1}{2}

(2n+1)π,23π+2nπ,43π+2nπ(2n+1)\pi, \frac{2}{3}\pi + 2n\pi, \frac{4}{3}\pi + 2n\pi

(8)

cos⁡2x+sin⁡2x=0cos⁡2x−sin⁡2x+sin⁡2x=0cos⁡2x=0x=π2+nπ\cos 2x + \sin^2 x = 0 \qquad \cos^2 x - \sin^2 x + \sin^2 x = 0 \qquad \cos^2 x = 0 \qquad x = \frac{\pi}{2} + n\pi

解説: r31bn1z