Find all solutions of the equation in interval [0, 2pi]

cos4x (cosx-1) =0

cos4x (cosx-1) =0

Either cos 4x = 0

OR cos x = 1
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cos 4x = 0

4x = pi/2, 3pi/2

x = pi/8, 3pi/8

x = 5pi/8, 7pi/8, 9pi/8, 11pi/8, 13pi/8, 15pi/8
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cos x = 1

x = 0, 2pi

Solutions: x = 0, pi/8, 3pi/8, 5pi/8, 7pi/8, 9pi/8, 11pi/8, 13pi/8, 15pi/8, 2pi
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here's a little hint about this part... DANIEL. You REALLY shouldn't copy others work.

cos(blah) = 0 when blah = pi/2 or 3pi/2 + 2pi*n

so cos(4x) = 0 when 4x = pi/2 or 3pi/2, solving for x we get

x = pi/8 + 2pi*n / 4 = pi/8 + pi*n/2
or
x = 3pi/8 + pi*n/2

the pi*n / 2 means I need to let n = 0, 1, 2... until we get values greater than 2pi and we'll stop

so

x = pi/8, pi/8 + 4pi/8 (common denom), pi/8 + 8pi/8, pi/8 + 12pi/8
here's a little hint about this part... cos(blah * x) will have blah number of answers
x = pi/8, 5pi/8, 9pi/8, and 13pi/8 and for the other one
x = 3pi/8, 7pi/8, 11pi/8, 15pi/8

For the other part, cos(x) - 1 = 0 or cos(x) = 1

cos(x) = 1 only at x = 0 + 2pi*n, we only need 2 answers here, 0 and 2pi since we are including both 0 and 2pi

So you have 10 solutions all together for this one.