Question

Solve each equation for 0 x 360 sqrt 2 cosx+1=0 please hurry

Answer 1

`120\º \text{ and } 240\º`

or in radians:

`$(2\pi )/(3) \text{ and } (4\pi )/(3)$`

Explanation:

From the way you wrote, you want to solve the equation

`\sqrt {2 \cos(x)+1}=0` for
`0 \leq x < 360\º`, or in radians
`[0, 2\pi)`

`\sqrt {2 \cos(x)+1}=0`

Square both sides

`2 \cos(x) +1= 0`

`2\cos(x)=-1`

`$\cos(x)=-(1)/(2) $`

In the Unit Circle, considering one revolution (interval
`[0, 2\pi)`),

the values where
`$\cos(x)=-(1)/(2) $` are in Quadrant II and III.

Once

`$\cos(x)= (1)/(2) \text{ for } x = 60\º \text{ in the Quadrant I}$`

The values where

`$\cos(x)=-(1)/(2) $` are 120º and 240º.

Answer 2

C on edge

Explanation: