Question

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

Answer 1

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