Respuesta :

Danutz98

Answer:

C.

Step-by-step explanation:

[tex]\cos x+\sqrt{2}=-\cos x \\ 2\cos x = -\sqrt 2 \\ \cos x = -\dfrac{\sqrt 2}{2}\\ \\ \Rightarrow x = \pm \arccos\Big(-\dfrac{\sqrt 2}{2}\Big)+2k\pi,\quad x\in \mathbb{Z}\\ \Rightarrow x =\pm\dfrac{3\pi}{4}+2k\pi\\ \\ \\k = -1 \Rightarrow x < 0\\\\ k=0 \Rightarrow x<0 \quad \text{or}\quad \boxed{x = \dfrac{3\pi}{4}} \\ \\k = 1 \Rightarrow x=-\dfrac{3\pi}{4}+2\pi \Rightarrow \boxed{x= \dfrac{5\pi}{4}}\quad \text{or}\quad x > 2\pi[/tex]

Answer:

C

Step-by-step explanation:

Given

cos x + [tex]\sqrt{2}[/tex] = - cosx ( add cosx to both sides )

2cosx +[tex]\sqrt{2}[/tex] = 0 (subtract [tex]\sqrt{2}[/tex] from both sides )

2cosx = - [tex]\sqrt{2}[/tex] ( divide both sides by 2 )

cosx = - [tex]\frac{\sqrt{2} }{2}[/tex]

Since cosx < 0 then x is in the second/ third quadrant

x = [tex]cos^{-1}[/tex] ( [tex]\frac{\sqrt{2} }{2}[/tex] )

  = [tex]\frac{\pi }{4}[/tex] ← related acute angle

Hence

x = π - [tex]\frac{\pi }{4}[/tex] = [tex]\frac{3\pi }{4}[/tex] ← second quadrant

or

x = π + [tex]\frac{\pi }{4}[/tex] = [tex]\frac{5\pi }{4}[/tex] ← third quadrant

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