Question

Resolva a equação 2cos2x−sinx−1=0 no intervalo 0⩽x⩽2π

Answer 1

Boa tarde

2*cos(2x) - sin(x) - 1 = 0

2*(cos²(x) - sen²(x)) - sen(x) - 1 = 0
2*(1 - 2sen²(x)) - sen(x) - 1 = 0 
2 - 4sen²(x)  - sen(x) - 1 = 0
4sen²(x) + sen(x) - 1 = 0 

y = sen²(x)

4y² + y - 1 = 0
d² = 1 + 8 = 9
d = 3

y1 = (-1 + 3)/8 = 1/4

x1 = 1/2, x = 30°, x = 150
x2 = -1/2 x = 210° ,, x = 330°