I factored the equation as follows:
Now, where we also have , so we need only concern ourselves with this, and for:
we find:
In other words, we want odd integer multiples of .
So there was an equation that I didn't quite understand, and I was hoping somebody could give me some insight. Here it goes...
Find ALL solutions to the equation: 3sintcost = 3cost
it was on mymathlab and the solution box was set up something like this: t = ____, where N is any integer.
the correct answer was: pi/2 + pi*N
my answer was: pi/2 + 2pi*N
*** I do not understand why it is "+ pi*N", because the sine and cosine functions have a periodicity of 2pi.
can someone explain why it is simply "+ pi*N" instead of "+ 2pi*N"??
I would like to add that the equation gets reduced to 3cos(t) ( sin(t) - 1)= 0
That implies either sin (t) = 1 or cos (t) = 0
We know that sin(t) = 1 when the terminal arm of angle t is at y axis. That has been rightly been explained that is t will be an odd multiple of pi/2
cos(t) = 0 when the terminal arm of angle t is at x axis that is when t is and even multiple of pi/2
now combining the two solutions we conclude that t = odd or even multiple of pi/2
and that means n * pi/2 when is any integer.