SOLUTION: Find the solutions in the interval [0, 2pi) 2 cos 3x = 1 This is due tomorrow and I am very confused, any and all help is greatly appreciated :)

Question 871008: Find the solutions in the interval [0, 2pi)
2 cos 3x = 1
This is due tomorrow and I am very confused, any and all help is greatly appreciated :)
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!

The cosine function takes all the positive values between 0 and 1, without repeating in quadrant I and again in quadrant IV.
In quadrants II and III cosine has negative values.
In quadrant I (between 0 and , or if you prefer between and ),
we have .
In quadrant IV , has a cosine of too.
Since cosine has a period of , adding you get an angle with the same cosine.
So all the angles with a cosine of can be expressed as
= , for any integer
If , then --> .
and are outside the interval

.
The solutions in the interval

are
, , , , , .
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