Question

Help please; I need the right answer ​

Answer 1

Answer:

C

Step-by-step explanation:

Given

cos x + $\sqrt{2}$ = - cosx ( add cosx to both sides )

2cosx +$\sqrt{2}$ = 0 (subtract $\sqrt{2}$ from both sides )

2cosx = - $\sqrt{2}$ ( divide both sides by 2 )

cosx = - $\frac{\sqrt{2} }{2}$

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

x = $cos^{-1}$ ( $\frac{\sqrt{2} }{2}$ )

  = $\frac{\pi }{4}$ ← related acute angle

Hence

x = π - $\frac{\pi }{4}$ = $\frac{3\pi }{4}$ ← second quadrant

or

x = π + $\frac{\pi }{4}$ = $\frac{5\pi }{4}$ ← third quadrant

Answer 2

Answer:

C.

Step-by-step explanation:

$\cos x+\sqrt{2}=-\cos x \\ 2\cos x = -\sqrt 2 \\ \cos x = -\dfrac{\sqrt 2}{2}\\ \\ \Rightarrow x = \pm \arccos\Big(-\dfrac{\sqrt 2}{2}\Big)+2k\pi,\quad x\in \mathbb{Z}\\ \Rightarrow x =\pm\dfrac{3\pi}{4}+2k\pi\\ \\ \\k = -1 \Rightarrow x < 0\\\\ k=0 \Rightarrow x 2\pi$