Sin⁡2x+cos⁡x=0 3sin⁡x+√3 cos⁡x=0 2sin^2 x+3sin⁡x-2=0 - вопрос №203302232022028862156 от tarasov78 20.04.2021 20:41

Question

Алгебра: Sin⁡2x+cos⁡x=0 3sin⁡x+√3 cos⁡x=0 2sin^2 x+3sin⁡x-2=0 sin^2 x-sin⁡x-2=0 2sin^2 x+sin⁡x cos⁡x-cos^2 x=0 √(16-x^2 )∙sin⁡x=0 sin⁡x+sin⁡2x=0 2cos^2 x-5cos⁡x+2=0 3sin^2 x-2 sin⁡x cos⁡x-cos^2 x=0 sin⁡x-cos⁡x=0 4sin^2 x-2 sin⁡x cos⁡x=3

tarasov78 · Answer

Sin⁡2x+cos⁡x=0 cos⁡x=-sin⁡2xcos⁡x=cos(-⁡2x-П/2)x = 3П/2 - 2x +2Пn1x = -3П/2 + 2x +2Пn2
x + 2x = 3П/2 + 2Пn13x = 3П/2 + 2Пn1x = 3П/2*3 + 2Пn1/3x = П/2 + 2Пn1/3
x - 2x= -3П/2 +2Пn2x -x= -3П/2 +2Пn2x= 3П/2 - 2Пn2