SOLUTION: Find all solutions of the equation {{{cos 3x = cos x}}} in the interval [0,2pi)

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Question 1152146: Find all solutions of the equation cos+3x+=+cos+x in the interval [0,2pi)
Found 2 solutions by MathLover1, math_helper:
Answer by MathLover1(20819) About Me  (Show Source):
You can put this solution on YOUR website!

Find all solutions of the equation cos%283x%29=cos%28x%29 in the interval [0,2pi)
cos%283x%29=cos%28x%29......write cos%283x%29 as cos%282x%2Bx%29 and use identity cos%28alpha+%2B+beta%29+=+cos%28alpha%29+cos%28beta%29+-+sin%28alpha%29+sin%28beta%29

cos%282x%2Bx%29=+cos%282x%29+cos%28x%29++-++sin%282x%29+sin%28x%29
now, write cos%282x%29 as cos%28x%2Bx%29 and use identity above
cos%28x%2Bx%29=+cos%28x%29+cos%28x%29++-++sin%28x%29+sin%28x%29

and sin%282x%29=sin%28x%2Bx%29=sin%28x%29+cos%28x%29+%2B+cos%28x%29+sin%28x%29

then we have


substitute in given equation:
cos%283x%29=cos%28x%29


....both sides divide by cos%28x%29
%28cos%5E2%28x%29++-sin%5E2%28x%29+%29++-+%282sin%5E2%28x%29+%29++=1
cos%5E2%28x%29++-sin%5E2%28x%29+++-+2sin%5E2%28x%29+++=1
cos%5E2%28x%29++-3sin%5E2%28x%29+++=1
++1-cos%5E2%28x%29%2B3sin%5E2%28x%29=0........use the identity: 1-cos+%5E2%28x%29=sin+%5E2%28x%29
sin+%5E2%28x%29%2B3sin%5E2%28x%29=0
4sin%5E2%28x%29=0
=> solutions:

if sin%5E2%28x%29=0 ...........apply rule: x%5En=0=>x=0
general solutions for sin%28x%29=0 are:
x=0%2B2pi%2An =>+x=2pi%2An
x=pi%2B2pi%2An
solutions for the range : 0%3C=x%3C2pi+
x=0, x=pi
combine solutions:
x=pi%2F2,+x=0, x=3pi%2F2,x=pi


Answer by math_helper(2460) About Me  (Show Source):
You can put this solution on YOUR website!
One way is to graph cos(x) and cos(3x) on [0,2pi) and see where the graphs have the same y-value:




Another way is to graph cos(x) - cos(3x) and look for points where this difference function crosses, or touches, the x-axis:





From the difference function, we see 0, pi/2, pi, and 3pi/2 are solutions.
Check:
cos(0) = cos(3*0) = 1
cos(pi/2) = cos(3pi/2) = 0
cos(pi) = cos(3pi) = -1
cos(3pi/2) = cos(9pi/2) = 0