Question

sen²x - ( 2senx.cosx - cos²x) = 0

 

a) 2Kpi - pi/6

b)kpi + pi/4

c) Kpi

d)2kpi + pi/6

e) kpi - pi/4

Answer 1

$\sin^2 x - \left ( 2 \cdot \sin x \cdot \cos x - \cos^2 x \right ) = 0 \\ \sin^2 x - 2 \cdot \sin x \cdot \cos x + \cos^2 x = 0 \\ \left (\sin^2 x + \cos^2 x \right ) - 2 \cdot \sin x \cdot \cos x = 0 \\ 1 - 2 \cdot \sin x \cdot \cos x = 0 \\ 2 \cdot \sin x \codt \cos x = 1 \\ \sin x = \frac{1}{2 \cdot \cos x}$

 

 Substituindo em: sen² x + cos² x = 1

 

 Teremos:

 

$\frac{1}{4 \cdot \cos^2 x} + \cos^2 x = 1 \\\\ 1 + 4 \cdot \cos^4 x = 4 \cdot \cos^2 x \\ 4 \cdot \cos^4 x - 4 \cdot \cos^2 x + 1 = 0 \\ (2 \cdot \cos^2 - 1)^2 = 0 \\ 2 \cdot \cos^2 x - 1 = 0 \\ \cos^2 x = \frac{1}{2} \\\\ \cos x = \pm \frac{\sqrt{2}}{2} \begin{cases} \cos x = \frac{\sqrt{2}}{2} \Rightarrow x = \frac{\pi}{4} \\ \cos x = - \frac{\sqrt{2}}{2} \Rightarrow x = \frac{3\pi}{4} \end{cases}$

 

  Logo,

 

$\boxed{\boxeed{S = \left \{ x \in \mathbb{R} / \frac{\pi }{4} + k\pi \right \}}}, k \in \mathbb{Z}$

 

 Daí, alternativa b.