SOLUTION: how do you find all solutions on [0,2pi) for the trig equation 2cos(3x)= -square root of 2 ?

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Question 1149401: how do you find all solutions on [0,2pi) for the trig equation
2cos(3x)= -square root of 2 ?
Answer by greenestamps(13195) About Me  (Show Source):
You can put this solution on YOUR website!


2cos%283x%29+=+-sqrt%282%29
cos%283x%29+=+-sqrt%282%29%2F2

-sqrt%282%29%2F2 is the cosine of an angle with a reference angle of 45 degrees (pi/4 radians) in quadrants II or III.

So the equation is satisfied whenever 3x is 3pi/4 or 5pi/4 radians, or either of those angles plus any multiple of 2pi.

3x = 3pi/4 --> x = pi/4
3x = 5pi/4 --> x = 5pi/12
3x = 3pi/4+2pi = 11pi/4 --> x = 11pi/12
3x = 5pi/4+2pi = 13pi/4 --> x = 13pi/12
3x = 3pi/4+4pi = 19pi/4 --> x = 19pi/12
3x = 5pi/4+4pi = 21pi/4 --> x = 21pi/12

Adding more multiples of 2pi will result in angles x outside the given range [0,2pi)

So 6 solutions....