# Math Help - sin(x)=cos(2x)

1. ## sin(x)=cos(2x)

Hi

Again I have to find the point of intersection of these two functions.
I know cos(2x) = cos(x)^2-sin(x)^2

But still: sin(x)=cos(x)^2-sin(x)^2, i have no idea how to solve that. I reached the point (1+sin(x))=tan(x)cos(x), but i don't get further. Anybody who can help to find the exact solution?

2. Originally Posted by Schdero
Hi

Again I have to find the point of intersection of these two functions.
I know cos(2x) = cos(x)^2-sin(x)^2

But still: sin(x)=cos(x)^2-sin(x)^2, i have no idea how to solve that. I reached the point (1+sin(x))=tan(x)cos(x), but i don't get further. Anybody who can help to find the exact solution?
hi

hint : cos 2x = 1-2sin^2 x

3. Originally Posted by Schdero
Hi

Again I have to find the point of intersection of these two functions.
I know cos(2x) = cos(x)^2-sin(x)^2

But still: sin(x)=cos(x)^2-sin(x)^2, i have no idea how to solve that. I reached the point (1+sin(x))=tan(x)cos(x), but i don't get further. Anybody who can help to find the exact solution?
$\cos^2(x) = 1-\sin^2(x)$

Consequently

$\cos(2x) = 1-2\sin^2(x)$

4. Originally Posted by Schdero
Hi

Again I have to find the point of intersection of these two functions.
I know cos(2x) = cos(x)^2-sin(x)^2

But still: sin(x)=cos(x)^2-sin(x)^2, i have no idea how to solve that. I reached the point (1+sin(x))=tan(x)cos(x), but i don't get further. Anybody who can help to find the exact solution?
Hi,

sin(x) = cos(x)^2-sin(x)^2
sin(x) = 1 - sin(x)^2 - sin(x)^2
2sin(x)^2 + sin(x) - 1 = 0
(2sin(x) - 1)(sin(x) + 1) = 0
sin(x) = -1/2 or sin(x) = 1

then solve for x.. =)

5. perfect, thanks!
I somehow simply don't connect the sin/cos rules well...
thanks again!

6. Originally Posted by Schdero
Hi

Again I have to find the point of intersection of these two functions.
I know cos(2x) = cos(x)^2-sin(x)^2

But still: sin(x)=cos(x)^2-sin(x)^2, i have no idea how to solve that. I reached the point (1+sin(x))=tan(x)cos(x), but i don't get further. Anybody who can help to find the exact solution?
Another approach is suggested by noting that sin(A) = cos(B) => sin(A) = sin(pi/2 - B).

Therefore A = pi/2 - B or A = pi - (pi/2 - B) ....