Find all the values of x in the interval [0,2π] that satisfy the equation: 8sin(2x)=8cos(x)
I think you mean x = π/2, π/6, 5π/6, 3π/2... 0 shouldnt be included
To find all the values of x in the interval [0, 2π] that satisfy the equation 8sin(2x) = 8cos(x), we can use trigonometric identities and solve for x step by step.
First, let's divide both sides of the equation by 8 to simplify it:
sin(2x) = cos(x)
Next, let's use the trigonometric identity sin(2x) = 2sin(x)cos(x), which gives us:
2sin(x)cos(x) = cos(x)
Now, let's rearrange the equation to isolate the cosine term:
2sin(x)cos(x) - cos(x) = 0
cos(x)(2sin(x) - 1) = 0
We now have two possible scenarios:
1) cos(x) = 0
2) 2sin(x) - 1 = 0
For case 1, when cos(x) = 0, it means x is either π/2 or 3π/2. These values fall within the interval [0, 2π].
For case 2, when 2sin(x) - 1 = 0, we can solve for x as follows:
2sin(x) - 1 = 0
2sin(x) = 1
sin(x) = 1/2
To find the values of x for which sin(x) = 1/2, we can refer to the unit circle or use trigonometric ratios. We know that sin(x) = 1/2 when x equals π/6 or 5π/6 in the first and second quadrants, respectively. These values also fall within the interval [0, 2π].
Therefore, the values of x that satisfy the equation 8sin(2x) = 8cos(x) in the interval [0, 2π] are:
x = π/2, 3π/2, π/6, 5π/6.
3 1/2+ 3/4 = xz\
^ that's not the answer
good catch anonymous.
that was just a typo, notice I actually had the right answer in the solution part.
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