sin4x = cos2x
i seem to have got up to thias point:
cos2x(2sin2x -1) = 0
but from here my answers to the book dont match up
it has one of the soltuions as kpi -pi/4
i think i have a 2 under the kpi. which is right?
its ok i think i realise now
the other soution doesn't quite match up to mine either. i got mine from 2sin2x-1 = 0
wot would be the general solution for this and why?
for cos2x(2sin2x -1) = 0
you not only have to consider when (2sin2x -1) = 0, but also if cos2x = 0 because either one could be zero to satisfy the equation, and its not nessacarily the second half
i get two solution for sumthing right.
when sinx = 1/2
x = 2npi + pi/6, 2npi + 5pi/6 - correct?
but my book seems to have kind of put these together to make 2npi/3 + pi/6
altho this seems to be for sinx = +/- 1/2
are you still considering that there is more than one value for sin2x=0? at sinx=0, x=0, pi, 2pi etc... so at sin2x=0, x=0, pi/2, pi etc... at cosx=0, x=pi/2, 3pi/2, 5pi/2... so at cos2x=0, x=pi/4, 3pi/4, 5pi/4... (or 1/2pi - pi/4, 1pi-pi/4, 3/2pi-pi/4) - which is your kpi-pi/4
Ive found the problem - the textbook i use makes use of the fact that sin X = cos(pi/2 - X)
simple i know - but it makes all the difference - YAYYYYYYYYYYYYY!!!!!!!!!!!!!!
what is the general solution for
(cosx + sinx)^2 = 3/2
last question on this exercise:
i have one solution but dont know how to get the other solution:
find the general solution of sqrt3cos2x + sin2x = 2sinx
cos(2x - pi/6) = cos(pi/2 - x)
(2x - pi/6) = (pi/2 - x)
3x = 2pi/3 + 2npi
x = 2pi/9 + 2npi/3
thats one correct soltuion
the other in the answers is 2npi - pi/3
how do i get this one?
I already mentioned how you could do
cosP = cosQ
equations by noting
cos(A+B) - cos(A-B) = -2sinAsinB
A+B = 2x -pi/6
A-B = pi/2 -x
actually because cos is even it's even easier than this
cosP = cosQ
P = Q + 2npi
P = -Q + 2npi
yep - have A = x/2 + pi/6 and B = -3x/2 + pi/3