Hi everybody,
I know, we are on a ASM forum but It's the end of holliday (here in France, in the south) but for Tuesday, I have to do a lot of homework and I 've not so much time to do it. It's in Mathematics :
Does someone know how I can solve the following:
cos²(x+pi/3)-sin²(2x+pi/4)=0
Thankx very much and sorry to ask this question here
Vom-bonjour:-()
PS:I did not manage to put the pi symbol.

what you have there is the difference of two perfect squares.
x^2 -y^2 = 0
(x-y)(x+y)=0
x is the cos term and y is the sin term.
rewrite cos(a) as sin(90-a) and solve x-y =0 and x+y =0.
For x-y=0, you will get sin(c) - sin(d) = 0
c=d ---> solve for x.
For x+y=0, you will get sin(c)+sin(d) = 0
therefore, c= 180-d . solve for x.
done!

***geesh***

Thankx very much, i didn't think to a²-b²=... Thankx
Vom-bonjour:-()

Hel, there are actually four solutions. sin(c) = sin(d) implies
c = d OR c = 180 - d. Also, sin(c) = - sin(d) implies c = 180 + d (not 180 - d) OR c = -d. Here c = 30 - x, d = 45 + 2x so there are four solutions for x.