# Thread: Trig equation problem using sum & difference identities

1. ## Trig equation problem using sum & difference identities

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

2. 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

3. Thanks for the post. How do i make (Cos 4x + sin 4x) = 0 ? i can't see how this would be done...

4. Originally Posted by dojo
Thanks for the post. How do i make (Cos 4x + sin 4x) = 0 ? i can't see how this would be done...
Express it as $r\sin(4x+\alpha)$.

5. 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