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

**darkmoon123** · April 14th 2009, 06:22 PM

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

**mr fantastic** · April 14th 2009, 06:49 PM

Originally Posted by darkmoon123

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

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

Solve these two equations over the given domain.

**darkmoon123** · April 14th 2009, 06:49 PM

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

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

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

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

$sinx=1$ $cosx=1$

{90 degrees, 0 degrees} ?

**mr fantastic** · April 14th 2009, 06:54 PM

Originally Posted by darkmoon123

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

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

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

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

$sinx=1$ $cosx=1$

{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?

**stapel** · April 15th 2009, 05:40 AM

Originally Posted by darkmoon123

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

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

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

$sinx=1$ $cosx=1$

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

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