Trigonometric equation trouble

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: 3**sin**t**cos**t = 3**cos**t

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"??

Re: Trigonometric equation trouble

I factored the equation as follows:

$\displaystyle 3\cos(t)(\sin(t)-1)=0$

Now, where $\displaystyle \sin(t)=1$ we also have $\displaystyle \cos(t)=0$, so we need only concern ourselves with this, and for:

$\displaystyle \cos(t)=0$

we find:

$\displaystyle t=\frac{\pi}{2}+N\pi=\frac{\pi}{2}(2N+1)$

In other words, we want odd integer multiples of $\displaystyle \frac{\pi}{2}$.

Re: Trigonometric equation trouble

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.

Re: Trigonometric equation trouble

$\displaystyle \cos(t)=0$ at odd multiples of $\displaystyle \frac{\pi}{2}$...at even multiples we have $\displaystyle |\cos(t)|=1$.

Re: Trigonometric equation trouble

Quote:

Originally Posted by

**MarkFL** $\displaystyle \cos(t)=0$ at odd multiples of $\displaystyle \frac{\pi}{2}$...at even multiples we have $\displaystyle |\cos(t)|=1$.

You are right i stand corrected. In both the cases t has to be odd multiple of pi/2