Posted by **Emily** on Thursday, September 13, 2012 at 7:40pm.

Find all the values of x in the interval [0,2π] that satisfy the equation: 8sin(2x)=8cos(x)

- math -
**Anonymous**, Thursday, September 13, 2012 at 7:42pm
3 1/2+ 3/4 = xz\

- math -
**Anonymous**, Thursday, September 13, 2012 at 7:48pm
^ thats not the answer

- math -
**Reiny**, Thursday, September 13, 2012 at 7:49pm
8sin(2x)=8cos(x)

8(2sinxcosx) - 8cosx = 0

16sinxcosx - 8cosx = 0

8cosx(2sinx - 1) = 0

cosx = 0 or sinx = 1/2

if cosx = 0, x = π/2 or x = 3π/2

if sinx = 1/2, x = π/6 or 5π/6

x = 0, π/6, 5π/6, 3π/2

- math -
**Anonymous**, Thursday, September 13, 2012 at 8:07pm
I think you mean x = π/2, π/6, 5π/6, 3π/2... 0 shouldnt be included

- you are correct - math -
**Reiny**, Thursday, September 13, 2012 at 8:14pm
good catch anonymous.

that was just a typo, notice I actually had the right answer in the solution part.

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