1. ## Trig equation

apply sum-to-product identity to solve equation for [0,360)
cos(t)-cos3(t)=0
the book says that the answer is 0,90,180,270,360,and 540.
I am getting 0,90,180,and 270, but thought that it was suppose to be less than 360.

sin(2t)-sin(5t)=0 [0,360)

i get -2cos(7t/2)sin(3t/2) and am totally lost on how to find the solutions.Do i divide 0 & 360 by 7/2 & 3/2 to find the solutions?The book has 0,pi/7,3pi/7,2pi/3,4pi/3, and 2pi.

Can anyone help me and explain what i am doing wrong?

2. Originally Posted by Thelema
apply sum-to-product identity to solve equation for [0,360)
cos(t)-cos3(t)=0
the book says that the answer is 0,90,180,270,360,and 540.
I am getting 0,90,180,and 270, but thought that it was suppose to be less than 360.
I think you are right

sin(2t)-sin(5t)=0 [0,360)

i get -2cos(7t/2)sin(3t/2) and am totally lost on how to find the solutions.Do i divide 0 & 360 by 7/2 & 3/2 to find the solutions?The book has 0,pi/7,3pi/7,2pi/3,4pi/3, and 2pi.

Can anyone help me and explain what i am doing wrong?
Did you know that the general solution to sin(x) = sin(y) is $x = n\pi + (-1)^n y$?(n is an integer)
[If you do not know the above think about it, its simple]

Then all you have to do now is sin(2t) = sin(5t)

So general solution is $2t = = n\pi + (-1)^n 5t$

$n = 0 \Rightarrow 2t = 5t \Rightarrow t = 0$

$n = 1 \Rightarrow 2t = \pi - 5t \Rightarrow t = \frac{\pi}7$

$n = 2 \Rightarrow 2t = 2\pi + 5t \Rightarrow t = -\frac{2\pi}3 \,\, or \,\, t = \frac{4\pi}{3}$

$n = 3 \Rightarrow 2t = 3\pi - 5t \Rightarrow t = \frac{3\pi}7$

Well I hope you get the idea

3. Thank you very much. I have not learned your method yet, how would i do it
0>7t/2>2pi?

4. Originally Posted by Thelema
Thank you very much. I have not learned your method yet, how would i do it
0>7t/2>2pi?
First observe that, for any trig function if x is a solution, then 2pi+x,4pi+x,6pi+x,...... is a solution.

Now observe that sin x = sin (pi - x). Which means if you get a solution with x as the answer, immediately (pi-x) qualifies too.

We can always add or subtract 2pi (that is 360 degrees OR one full rotation) to any angle.So in summary, if x is a solution, then (pi-x), (2pi+x), (3pi-x), (4pi+x),..... are all solutions.Look at the pattern, and guess the general form.

If x is a solution for a sin equation, then the generalised form solution is $n\pi + (-1)^n x$

The best part about this is that you have to just obtain one solution in $[0, \frac{\pi}2]$, Then you can plug in that solution as x in the above formula to get all possible solutions.
That is exactly what I did in the previous post with 5t and 2t.
Got it?