1-sin2x = sinx-cosx
sin²x+cos²x-2·sin x·cos x = sin x - cos x
(sin x - cos x)² = sin x - cos x
(sin x - cos x)(sin x - cos x - 1) = 0
1) sin x - cos x = 0
tg x = 1
2) sin x - cos x - 1 = 0
sin x - cos x = 1
ответ:
1-sin2x = sinx-cosx
sin²x+cos²x-2·sin x·cos x = sin x - cos x
(sin x - cos x)² = sin x - cos x
(sin x - cos x)(sin x - cos x - 1) = 0
1) sin x - cos x = 0
tg x = 1
2) sin x - cos x - 1 = 0
sin x - cos x = 1
ответ:![\frac{\pi}{4}+\pi k;\ -\frac{11\pi}{12}+2\pi n;\ \frac{5\pi}{12}+2\pi n; \ k,n \in Z](/tpl/images/1356/2556/767e6.png)