Re: Finding angles of intersection

On Aug 9, 11:59 am, "Konrad Viltersten" <t...@xxxxxxxxxxxxxx> wrote:
Find x so that
sin(4x) == -cos(2x)

or sin(4x) = -cos(-2x) since cos(a) = cos(-a)

I perform:
sin(4x) == -sin(2x-a), a=pi/2

This is wrong. Should be

sin(4x) = -sin(a-2x)

But noting what I wrote above

sin(4x) = -sin(a+-2x)

sin(4x) == sin(-2x+a)
4x == -2x + a +2pi*n
6x == a + 2pi*n
x == pi/12 + n*pi/3

Following your technique:

4x = -a +-2x + 2pi*n

6x = (2n - 0.5)*pi
or
2x = (2n - 0.5)*pi

x = (n/3 - 1/12)*pi = (4n-1)*pi/12
or
x = (n - 1/4)*pi = (2n-1)*pi/4

So values in the range [0,pi] are:
{pi/4, 7pi/12, 3pi/4, 11pi/12} or
{3pi/12, 7pi/12, 9pi/12, 11pi/12}


Hence i get solutions:
pi/12 5pi/12 9pi/12

Try these solutions, two of them don't solve your
original equation.

- Randy

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