Question

Help please; I need the right answer ​

Answer 1

C

Explanation:

Given

cos x +
`√(2)` = - cosx ( add cosx to both sides )

2cosx +
`√(2)` = 0 (subtract
`√(2)` from both sides )

2cosx = -
`√(2)` ( divide both sides by 2 )

cosx = -
`(√(2) )/(2)`

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

x =
`cos^(-1)` (
`(√(2) )/(2)` )

=
`(\pi )/(4)` ← related acute angle

Hence

x = π -
`(\pi )/(4)` =
`(3\pi )/(4)` ← second quadrant

or

x = π +
`(\pi )/(4)` =
`(5\pi )/(4)` ← third quadrant

Answer 2

C.

Explanation:

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