Question

Please help me understand solving the following Trig Equation: 2cos3x=1

Answer 1

\[Interval: 0 \le x \le 2\pi \]

Answer 2

to start with the problem, divide both sides by 2.

Answer 3

Yes I did start that way

Answer 4

\[\large \cos 3x = \frac{ 1 }{ 2 }\]

Answer 5

how would I get rid of that 3 is the concern

Answer 6

since the required angle x is \[\large 0 \le x \le 2 \pi\] we multiply this by 3 to get the range of angle measures as solution\[\large 0 \le 3x \le 6 \pi\]

Answer 7

what angle (3x) gives \[\large (3x) = 1/2\space \text ?\]

Answer 8

hmmm....would there be any way to get that 3 to the other side? could I divide it from both sides? or is that even necessary?

Answer 9

let's rephrase the question. what angle y will satisfy \[\large \cos y = 1/2\]

Answer 10

\[\pi/3 ?\]

Answer 11

sorry for the typo ^. it is \[\large \cos (3x)=1/2\]

Answer 12

yes, pi/3 is one. can you give another?

Answer 13

\[5\pi/3?\]

Answer 14

great. these are the answers when the angle range is \[\large 0 \le y \le 2\pi\] what about the intervals \[\large 2\pi \le y \le 4\pi\]and \[\large 4\pi \le y \le 6\pi\]?

Answer 15

Let me just get one thing clear. When you multiplied the interval by 3...that in turn made is so the 3 in the problem didnt matter?

Answer 16

Im not to clear on how that worked

Answer 17

just add 2pi and 4pi to each of your answer. like pi/3 + 2pi, 5pi/3 + 2pi, pi/3 + 4pi, 5pi/3 + 4pi

Answer 18

the possible solutions are \[\large (3x) = \pi/3, 5\pi/3, 7\pi/3, 11\pi/3, 13\pi/3, 17\pi/3\]

Answer 19

because we're not used to solving cos (3x) when 0 < x < 2pi. the coefficients of x don't seem to be in agreement. so it is necessary to multiply by 3 just to make these coefficients the same.

Answer 20

ok so if it was say 4 instead of 3 you would multiply the interval by 4

Answer 21

right. this way, you get to know the angle solutions. so in your problem, 0 < 3x < 6pi, the angles to report are pi/3, 5pi/3, 7pi/3, 11pi/3, 13pi/3, 17pi/3. all these angles are less than 6pi.

Answer 22

\[\large 0\le y \le 2\pi:\pi/3, 5\pi/3\\\large 2\pi \le y \le 4\pi: 7\pi/3, 11\pi/3\\\large 4\pi \le y \le 6\pi: 13\pi/3,17\pi/3\]

Answer 23

ok i see

Answer 24

\[\large 3x=\pi/3,5\pi/3,7\pi/3,11\pi/3,13\pi/3,17\pi/3\]just divide everything by 3
\[\large x=\pi/9,5\pi/9,7\pi/9,11\pi/9,13\pi/9,17\pi/9\]

Answer 25

observe that the largest angle 17pi/9, is still less than 2pi as required in the problem \[\large 0 \le x \le 2\pi\]