In Mathematics could someone please explain this: 2cos6xcos2x=cos6x is the same as, cos2x=21 or cos6x=0?

Question

this is trigonometry, this: 2cos6xcos2x=cos6x is supposedly equal to, cos2x=21 or cos6x=0 but I don't understand how this was reached.

Answer 1

Move everything to one side, giving you

2 * cos6x * cos2x - cos6x = 0

then factor out cos6x to get

(cos6x) * (2cos2x - 1 ) = 0.

So either cos6x = 0 or 2cos2x - 1 = 0 -> cos2x = 1/2.

cos6x = 0 when 6x has the form ((2k + 1)/2) * pi for every integer k, so x will have the form ((2k + 1)/12) * pi, again for any integer k.

cox2x = 1/2 when 2x has the form (3h + 1)/3 * pi for every integer h, so in this case x has the form (3h + 1)/6 * pi for every integer h.

Thus, your answer is when x has the forms:

(2m + 1)/12 * pi or (3m + 1)/6 * pi for every integer m; there are infinitely many solutions.

Answer 2

Is 2cos6xcos2x=cos6x the same as, cos2x=21 or cos6x=0?

There are some problems here.


First cos(2x) can never be 21. The max it can get is 1.


Second: Solving the equation
2cos6xcos2x=cos6x

Multiply both sides by sec(6x) .. same as 1/cos(6x)
2cos(2x)=1

Multiplying both sides by 1/2
cos(2x)=1/2
So, cos(2x) = 1/2, not 21 (First problem)

x=π/6, I happen to know that cos(π/3)=1/2
So cos(6x)=cos(π) = -1, not 0


Third: Interestingly, had the problem been 2sin(6x)sin(2x)=sin(6x)
Then x would have been π/12,
sin(2x) = sin(2π/12) = sin(π/6) = 1/2
sin(6x) = sin(6π/12) = sin(π/2) = 0

Answer 3

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Answer 4

2cos6x cos2x=cos6x
cos6x=0,6x=pi/2.3pi/2
ORcos2x=1/2=cospi/3,
2x=pi/3,5pi/3