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anonymous

  • one year ago

Solve the equation algebraically: tanxsinxcosx - 1 = 0 Now, can someone explain why the answer has no solution? Haalllppppp

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  1. anonymous
    • one year ago
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    perhaps if we rewrite \[\tan(x)\cos(x)\sin(x)\] it will be more clear

  2. anonymous
    • one year ago
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    since \[\tan(x)=\frac{\sin(x)}{\cos(x)}\] you are looking at \[\frac{\sin(x)}{\cos(x)}\sin(x)\cos(x)=\sin^2(x)\]

  3. anonymous
    • one year ago
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    solving \[\sin^2(x)-1=0\] is not hard what do you get?

  4. anonymous
    • one year ago
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    So we don't factor it like \[(\sin(x)-1) (\sin(x)+1) = 0?\]

  5. anonymous
    • one year ago
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    you can if you like

  6. anonymous
    • one year ago
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    or you can just write \[\sin^2(x)=1\]so \[\sin(x)=1\] or \[\sin(x)=-1\]

  7. anonymous
    • one year ago
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    idk apparently the answer doesn't have a solution, that's where i got confused :/

  8. anonymous
    • one year ago
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    we are not done yet!

  9. anonymous
    • one year ago
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    LOL

  10. anonymous
    • one year ago
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    once we solve \[\sin(x)=1\] for \(x\) we can see why there is no solution did you solve it yet?

  11. anonymous
    • one year ago
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    How so? I thought the solution can also be \[\pi/2\] or \[3\pi/2\]

  12. anonymous
    • one year ago
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    so that makes it undefined?

  13. anonymous
    • one year ago
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    right

  14. anonymous
    • one year ago
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    since at those number, cosine is zero

  15. anonymous
    • one year ago
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    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

  16. anonymous
    • one year ago
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    ahah, that makes a lot of sense now hehe :) tysm!

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