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

Answer 1

3 1/2+ 3/4 = xz\

Answer 2

^ that's not the answer

Answer 3

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

Answer 4

I think you mean x = π/2, π/6, 5π/6, 3π/2... 0 shouldnt be included

Answer 5

good catch anonymous.

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