Question

Resolva a equação LaTeX: 2\cos^2x - \sin x - 1 = 02cos2x−sinx−1=0 no intervalo LaTeX: 0 \leqslant x \leqslant 2\pi

Answer 1

2cos² x - sen x - 1 = 0
2(1 - sen² x) - sen x - 1 = 0
2 - 2sen² x - sen x - 1 = 0
- 2sen² x - sen x + 1 = 0
2sen² x + sen x - 1 = 0

sen x' = 2

sen x" = - 5/2

cos 2x - sen x - 1 = 0
cos² x - sen² x - sen x - 1 = 0
1 - sen² x - sen² x - sen x - 1 = 0
- 2sen² x - sen x = 0
2sen² x + sen x = 0
sen x(2sen x + 1) = 0

2sen x + 1 = 0
2sen x = - 1
sen x = - 1/2

S = {π/6, 5π/6)