anonymous
  • anonymous
How do you solve for sinx=cosx [0,2pi] ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\sin x=\cos x\\ \frac{\sin x}{\cos x}=\frac{\cos x}{\cos x}\\ \tan x=1\] When is tangent equal to 1?
anonymous
  • anonymous
that would be at pi/4 and 5pi/4...right?
anonymous
  • anonymous
which is kind of like asking "where is sine equal to cosine"

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anonymous
  • anonymous
Yes
anonymous
  • anonymous
OH!!!
anonymous
  • anonymous
funny how restating a problem makes it easier right?
DebbieG
  • DebbieG
Just remember... it isn't always "safe" to divide by something like cos(x). But you can get away with it here, since the solution set requires that sin(x)=cos(x) and that never happens anywhere that cos(x)=0. :)
anonymous
  • anonymous
yeah LOL oh, and how do I find the period of the following trig equations? y=-3tan pi x -3sec(- 6x) +2
DebbieG
  • DebbieG
Just don't try it if you have: \(\Large \cos^2(x)=cos(x)\) :)
anonymous
  • anonymous
try it?
DebbieG
  • DebbieG
dividing by cos(x)
anonymous
  • anonymous
try what? Is that impossibl?
anonymous
  • anonymous
*impossible to solve?
anonymous
  • anonymous
why is it bad to do that?
anonymous
  • anonymous
the period of \[\tan(bx)\] is \(\frac{\pi}{b}\)
DebbieG
  • DebbieG
If B is the coefficient on x (e.g., if you have y=sin(2x) then B=2) then: For sine, cosine, cosecant and secant, period = 2pi/B For tangent and cotangent, period = pi/B
DebbieG
  • DebbieG
Sorry, @lucy4104 , did you read my first comment above also? I didn't mean to confuse you, was just saying that you can't ALWAYS go dividing an equation by an unknown (like cos(x)) but it was OK here. In a different equation, you can't do it!
anonymous
  • anonymous
why not...?
DebbieG
  • DebbieG
Because what if cos(x)=0? :)
DebbieG
  • DebbieG
\[\Large \cos^2(x)=\cos(x)\]\[\Large \cos^2(x)-\cos(x)=0\]\[\Large \cos(x)(\cos(x)-1)=0\] So cos(x)=0, hence x=pi/2 or 3pi/2.... OR cos(x)=1, hence x=0 That's 3 solutions, but if you start THIS one by dividing by cos(x), you end up losing the first 2 solutions, because you eliminate the cos(x)=0 part of the solution.
anonymous
  • anonymous
cos(x)=0... then x is pi/2 +n(pi)
DebbieG
  • DebbieG
But it was OK to do so for YOUR equation above, since when cos(x)=sin(x), cos(x) is never = 0.
DebbieG
  • DebbieG
Yes, if you want the general form solution set with ALL real numbers.
anonymous
  • anonymous
oh, ok then you're saying to watch out? Wait, then how can you tell if its ok?
DebbieG
  • DebbieG
In general, don't divide by an unknown if it COULD be =0. But in your original equation cos(x)=sin(x) the solution set can't include any values where cos(x)=0, since that is true (x=pi/2 and x=3pi/2 and all coterminals) ONLY at angles where cos(x) is NOT = sin(x). So you can divide by cos(x) because it can't =0. Don't get too hung up on it. :) it was kind of an aside. :)
DebbieG
  • DebbieG
It's just like in algebra, if you solve \(x^2=x\) you do it by setting = 0 and factoring, not by dividing by x. Same idea.
anonymous
  • anonymous
oh, ok now i c it~ That was a good way to put it. Thank you!

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