anonymous
  • anonymous
Solve the equation algebraically: tanxsinxcosx - 1 = 0 Now, can someone explain why the answer has no solution? Haalllppppp
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
perhaps if we rewrite \[\tan(x)\cos(x)\sin(x)\] it will be more clear
anonymous
  • anonymous
since \[\tan(x)=\frac{\sin(x)}{\cos(x)}\] you are looking at \[\frac{\sin(x)}{\cos(x)}\sin(x)\cos(x)=\sin^2(x)\]
anonymous
  • anonymous
solving \[\sin^2(x)-1=0\] is not hard what do you get?

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anonymous
  • anonymous
So we don't factor it like \[(\sin(x)-1) (\sin(x)+1) = 0?\]
anonymous
  • anonymous
you can if you like
anonymous
  • anonymous
or you can just write \[\sin^2(x)=1\]so \[\sin(x)=1\] or \[\sin(x)=-1\]
anonymous
  • anonymous
idk apparently the answer doesn't have a solution, that's where i got confused :/
anonymous
  • anonymous
we are not done yet!
anonymous
  • anonymous
LOL
anonymous
  • anonymous
once we solve \[\sin(x)=1\] for \(x\) we can see why there is no solution did you solve it yet?
anonymous
  • anonymous
How so? I thought the solution can also be \[\pi/2\] or \[3\pi/2\]
anonymous
  • anonymous
so that makes it undefined?
anonymous
  • anonymous
right
anonymous
  • anonymous
since at those number, cosine is zero
anonymous
  • anonymous
don't forget you stared with \[\tan(x)\sin(x)\cos(x)\] and since cosine is zero at those number, a) tangent is not defined there and b) cosine is zero, so you would actually have \(\frac{1}{0}\times 1\times 0\) moral of the story you cannot cancel zeros
anonymous
  • anonymous
ahah, that makes a lot of sense now hehe :) tysm!

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