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GaryK [48]
3 years ago
12

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

Mathematics
1 answer:
jolli1 [7]3 years ago
7 0

Answer:

120\º \text{ and } 240\º

or in radians:

$\frac{2\pi }{3} \text{ and } \frac{4\pi }{3}$

Step-by-step 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

<u>Square both sides</u>

2 \cos(x) +1= 0

2\cos(x)=-1

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

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

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

Once

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

The values where

$\cos(x)=-\frac{1}{2} $   are 120º and 240º.

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