# help: solving equation (cos^2x)(sin^2x)=0

• Apr 14th 2009, 06:22 PM
darkmoon123
help: solving equation (cos^2x)(sin^2x)=0
i need some help resolvong $\displaystyle \cos^2 x \sin^2 x = 0$ over the interval [0 degrees and 360 degrees]
• Apr 14th 2009, 06:49 PM
mr fantastic
Quote:

Originally Posted by darkmoon123
i need some help resolvong $\displaystyle \cos^2 x \sin^2 x = 0$ over the interval [0 degrees and 360 degrees]

So either $\displaystyle \sin x = 0$ or $\displaystyle \cos x = 0$.

Solve these two equations over the given domain.
• Apr 14th 2009, 06:49 PM
darkmoon123
I think i have the answer but i am not sure, can someone check this:

$\displaystyle (cos^2x)(sin^2x)=0$

$\displaystyle (1-sin^2x ) = 0$ $\displaystyle (1-cos^2x) = 0$

$\displaystyle sin^2x=1$ square root both sides $\displaystyle cos^2x=1$ square root both sides

$\displaystyle sinx=1$ $\displaystyle cosx=1$

{90 degrees, 0 degrees} ?
• Apr 14th 2009, 06:54 PM
mr fantastic
Quote:

Originally Posted by darkmoon123
I think i have the answer but i am not sure, can someone check this:

$\displaystyle (cos^2x)(sin^2x)=0$

$\displaystyle (1-sin^2x ) = 0$ $\displaystyle (1-cos^2x) = 0$

$\displaystyle sin^2x=1$ square root both sides $\displaystyle cos^2x=1$ square root both sides

$\displaystyle sinx=1$ $\displaystyle cosx=1$

{90 degrees, 0 degrees} ?

Did you read what I posted?

And you have only found solutions over [0 degrees, 90 degrees]. And what about the other solutions?
• Apr 15th 2009, 05:40 AM
stapel
Quote:

Originally Posted by darkmoon123
$\displaystyle (cos^2x)(sin^2x)=0$

$\displaystyle (1-sin^2x ) = 0$ $\displaystyle (1-cos^2x) = 0$

$\displaystyle sin^2x=1$ square root both sides $\displaystyle cos^2x=1$ square root both sides

$\displaystyle sinx=1$ $\displaystyle cosx=1$

If you want to use this method, instead of the one explained earlier, you'll need to use what you learned back in algebra about solving quadratic equations.

In particular, if $\displaystyle 1\, -\, \sin^2(x)\, =\, 0$, then $\displaystyle (1\, -\, \sin(x))(1\, +\, \sin(x))\, =\, 0$, so $\displaystyle \sin(x)\, =\, \pm 1$. And so forth.

(Wink)