Hi
(i) correct
(ii) Cos 5x + Sin 5x + Cos 3x + Sin 3x = 0
Cos 5x + Cos 3x + Sin 5x + Sin 3x = 0
2 Cos 4x * Cos x + 2 Sin 4x * Cos x = 0
2 Cos x (Cos 4x + Sin 4x) = 0
i let you finish the job
Hi there,
I'm looking for some help on two questions, I think I have the first one but would like confirmation that I'm correct. The second one I'm getting slightly confused with.
thanks.
Solve the following for 0 <= x <= pi
(i) Sin 5x - sin 2x = 0
(ii) Cos 5x + Sin 5x + Cos 3x + Sin 3x = 0
Okay,
For (i):
I use the sum and difference identities and get:
2cos (7x/2) * sin (3x/2) = 0
Therefore when either of these equal zero the statement is true.
Therefore,
2cos (7x/2) = 0
x = pi/2 as a general solution and calculate the next six.
the divide by 7/2 to give x and select the ones in the correct range which are:
pi/7, 3pi/7, 5pi/7 & pi
Do the same for Sine (3x/2) = 0 and I get 0 & 2pi/3.
These are all the solutions.
Is this correct??
(ii) It looks like this is a case of two sum and difference identities combined but I just get into a mess.
Can anyone help???
many thanks,
D
OK, so I get:
... 2 Cos x (Cos 4x + Sin 4x) = 0
2 cos x = 0 therefore x = pi/2
cos 4x + sin 4x = 0
Therefore, r sin( 4x - alpha) = 0
sin (4x - pi/4) = 0
(4x - pi/4) = 0, pi, 2*pi, 3*pi ...
4x = pi/4 , 5pi/4 , 9pi/4 , 13pi/4
x = pi/16 , 5pi/16 , 9pi/16 , 13pi/16
So as the range is 0 <= x <= pi
the full answers are:
pi/16 , 5pi/16 , 9pi/16 , 13pi/16 , pi/2
Is this correct??
thanks for all your help
D