Question

Por favor podem me resolver esta equação? $2 cos ^{2} x+sinx-1=0$

Answer 1

Olá Rgs

2cos²(x) + sen(x) - 1 = 0

cos²(x) + sen²(x) = 1
cos²(x) = 1 - sen²(x)

2 - 2sen²(x) + sen(x) - 1 = 0

2sen²(x) - sen(x) - 1 = 0

(sen(x) - 1)*(1 + 2sen(x) = 0 

sen(x) - 1 = 0
sen(x) = 1

x = π/2 + 2kπ

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

x = 7π/6 + 2kπ
x = 11π/6 + 2kπ
x = π/2 + 2kπ 

Answer 2

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

Veja o anexo
sen x = -1/2                   e      sen x = 1, logo
x = 7π/6 + 2kπ                 x = π/2 + 2kπ
x = -π/6 + 2kπ 

$S = \left\{x\in\Re\mid x= \dfrac{\pi}{2}+2k\pi \ ou\ x= -\dfrac{\pi}{6}+2k\pi \ ou \ x= \dfrac{7\pi}{6}+2k\pi \right\}$