How do i solve for x given, 2sinxcosx = cos2x?

Question

I've tried it so many times...please help!

Answer 1

2sinxcosx = cos2x

Your first step would be to note the identity
2sinycosy = sin2y

We apply this identity to the left hand side, to get

sin(2x) = cos(2x)

Now, we divide both sides by cos(2x), to get

sin(2x) / cos(2x) = 1

Note that sine over cosine is the definition of tan.

tan(2x) = 1

And now we solve this normally. Assuming we have a restricted domain of 0 <= x < 2pi,

2x = {pi/4, 5pi/4, 9pi/4, 13pi/4}

Note that I added the two values 9pi/4 and 13pi/4 knowing that they're well over the restriction of 2pi. However, if we divide both sides by 2, we get

x = {pi/8, 5pi/8, 9pi/8, 13pi/8}

Answer 2

2sinxcosx = cos2x

sin2x = cos2x = sin(pi/2 -2x) = sin(pi/2 -2x + 2npi)

2x = pi/2 -2x + 2npi

x = (1/4)(pi/2 + 2npi) = pi/8 + n pi/2, n is any integer.

Answer 3

Well, 2sinxcosx is the identity for sin2x.

So you've got

sin2x = cos2x

Divide both sides by cos2x and you get:

tan2x = 1

Tangent is 1 at π/4 + kπ. (This means that tangent is 1 at π/4, π/4 + π = 3π/4, π/4 + 2π = 5π/4, ...)

So:

2x = π/4 + kπ
x = π/8 + kπ/2

(π/8, π/8 + π/2 = 5π/8, π/8 + π = 9π/8....)

Answer 4

First convert as follows this is a known formula:
2sin(x)cos(x) = sin(2x)
So, than your equation is sin(2x) = cos(2x)
That means that 2x = 45 degrees
so x = 22.5 degrees.

If you want you can solve it radians then
2x = pi/4
x= pi/8.

Answer 5

sin2x=cos2x sin(pi/4)=cos(pi/4) so 2x=pi/4 x= pi/8 + kPi/2