Question

factor

Answer 1

\[2\cos^2x-1-cosx=0\]

Answer 2

\[(2\cos(x)+1)(\cos(x)-1)=0\]
All good?

Answer 3

ohhh okay yeah thanks

Answer 4

so it's 1/2 and 1

Answer 5

is that correct?

Answer 6

no yeah we have to solve it

Answer 7

\[(2\cos(x)+1)(\cos(x)−1)=0 \\ \\ 2\cos(x)+1=0 ~~~~~~ \cos(x)-1=0 \\ \\ ~~~~~~ ? ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ?\]

Answer 8

oops it's -1/2 and 1

Answer 9

right?

Answer 10

Yep now find 'x' by inversing cosine

Answer 11

I don't think i need to do that for this problem because the directions say to solve the equation for \[0lex <2\pi\]

Answer 12

sorry le is supposed to be less than or equal to

Answer 13

The question is asking you to find the angles in radians.

Answer 14

So you have
\[\cos(x)=-\frac{1}{2} ~~~~~~ and ~~~~~~ \cos(x)=1 \\ \\ x=\cos^{-1}(-\frac{1}{2}) ~~~~~~ and ~~~~~~ x=\cos^{-1}(1)\]

Answer 15

First, ignore the negative sign,
\[ x=\cos^{-1}(-\frac{1}{2}) \\ \\ x=\frac{\pi}{3}\]
For cosine with a negative number inside the inverse cosine function, you will need to add and subtract \(\pi\) to get two values/answers.
\[\pi + \frac{\pi}{3} = \frac{4}{3} \pi\]
\[\pi - \frac{\pi}{3} = \frac{2}{3} \pi\]
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Now for \[x=\cos^{-1}(1) \\ \\ x=0\]
You needn't have to add and subtract because the number is positive inside the inverse cosine.
So your final answers are:
\[x=0, \frac{2}{3} \pi, \frac{4}{3} \pi ~~~,~~~ 0 < x < 2\pi\]

Answer 16

Not zero, sorry because it is out of range.
\[x=\frac{2}{3} \pi, \frac{4}{3} \pi ~~~,~~~ 0 < x < 2\pi\]

Answer 17

okay i get you now

Answer 18

i totally forgot about the unit circle part of it

Answer 19

thanks for helping me!