Why the result is not included?

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If i have this exercise:



$tan(alpha) = 0$, the solution is when the $sin(a) = 0$, that is in $0°$ and $180°$, because $tan = sin/cos$



So the solution must be: $alpha_1 = 0 + pi k$, $alpha_2 = pi + pi k$



But according to symbolab, the unique solution is the $alpha_1$, so why $alpha_2$ is incorrect?




trigonometry




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  • Check the answer again: It should be just $pi k$, since $tan$ has a period of $pi$.
    – quasi
    Aug 23 at 2:15











  • also it is bad form to mix degrees and radians $180 = pi$.
    – Andrew Allen
    Aug 23 at 2:17










  • I already edited, but why the answer $pi + pi k$ isn't included?
    – Mattiu
    Aug 23 at 2:18






  • 3




    The set of solutions described by your $alpha_1$ are $dots,picdot(-1),picdot 0, picdot 1, picdot 2,picdot 3,dots$. The set of solutions described by your $alpha_2$ are $dots,pi+picdot(-1),pi+picdot 0,pi+picdot 1,pi+picdot 2,dots$ which you notice after simplification is exactly the same set as the first. That doesn't make it "incorrect", just redundant.
    – JMoravitz
    Aug 23 at 2:21







  • 2




    The answer is neither $0+pi k$ nor $pi+pi k,$ but , as in the previous comment by dxiv, the answer is "$tan a=0 iff ain pi k:kin Bbb Z."$...... $pi k$ is a number. The solution is not a number. It is an infinite set of numbers.
    – DanielWainfleet
    Aug 23 at 5:46















up vote
-1
down vote

favorite












If i have this exercise:



$tan(alpha) = 0$, the solution is when the $sin(a) = 0$, that is in $0°$ and $180°$, because $tan = sin/cos$



So the solution must be: $alpha_1 = 0 + pi k$, $alpha_2 = pi + pi k$



But according to symbolab, the unique solution is the $alpha_1$, so why $alpha_2$ is incorrect?




trigonometry




share|cite|improve this question






















  • Check the answer again: It should be just $pi k$, since $tan$ has a period of $pi$.
    – quasi
    Aug 23 at 2:15











  • also it is bad form to mix degrees and radians $180 = pi$.
    – Andrew Allen
    Aug 23 at 2:17










  • I already edited, but why the answer $pi + pi k$ isn't included?
    – Mattiu
    Aug 23 at 2:18






  • 3




    The set of solutions described by your $alpha_1$ are $dots,picdot(-1),picdot 0, picdot 1, picdot 2,picdot 3,dots$. The set of solutions described by your $alpha_2$ are $dots,pi+picdot(-1),pi+picdot 0,pi+picdot 1,pi+picdot 2,dots$ which you notice after simplification is exactly the same set as the first. That doesn't make it "incorrect", just redundant.
    – JMoravitz
    Aug 23 at 2:21







  • 2




    The answer is neither $0+pi k$ nor $pi+pi k,$ but , as in the previous comment by dxiv, the answer is "$tan a=0 iff ain pi k:kin Bbb Z."$...... $pi k$ is a number. The solution is not a number. It is an infinite set of numbers.
    – DanielWainfleet
    Aug 23 at 5:46













up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











If i have this exercise:



$tan(alpha) = 0$, the solution is when the $sin(a) = 0$, that is in $0°$ and $180°$, because $tan = sin/cos$



So the solution must be: $alpha_1 = 0 + pi k$, $alpha_2 = pi + pi k$



But according to symbolab, the unique solution is the $alpha_1$, so why $alpha_2$ is incorrect?




trigonometry




share|cite|improve this question














If i have this exercise:



$tan(alpha) = 0$, the solution is when the $sin(a) = 0$, that is in $0°$ and $180°$, because $tan = sin/cos$



So the solution must be: $alpha_1 = 0 + pi k$, $alpha_2 = pi + pi k$



But according to symbolab, the unique solution is the $alpha_1$, so why $alpha_2$ is incorrect?





trigonometry





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 23 at 7:04









N. F. Taussig

38.8k93153




38.8k93153










asked Aug 23 at 2:11









Mattiu

727317




727317











  • Check the answer again: It should be just $pi k$, since $tan$ has a period of $pi$.
    – quasi
    Aug 23 at 2:15











  • also it is bad form to mix degrees and radians $180 = pi$.
    – Andrew Allen
    Aug 23 at 2:17










  • I already edited, but why the answer $pi + pi k$ isn't included?
    – Mattiu
    Aug 23 at 2:18






  • 3




    The set of solutions described by your $alpha_1$ are $dots,picdot(-1),picdot 0, picdot 1, picdot 2,picdot 3,dots$. The set of solutions described by your $alpha_2$ are $dots,pi+picdot(-1),pi+picdot 0,pi+picdot 1,pi+picdot 2,dots$ which you notice after simplification is exactly the same set as the first. That doesn't make it "incorrect", just redundant.
    – JMoravitz
    Aug 23 at 2:21







  • 2




    The answer is neither $0+pi k$ nor $pi+pi k,$ but , as in the previous comment by dxiv, the answer is "$tan a=0 iff ain pi k:kin Bbb Z."$...... $pi k$ is a number. The solution is not a number. It is an infinite set of numbers.
    – DanielWainfleet
    Aug 23 at 5:46

















  • Check the answer again: It should be just $pi k$, since $tan$ has a period of $pi$.
    – quasi
    Aug 23 at 2:15











  • also it is bad form to mix degrees and radians $180 = pi$.
    – Andrew Allen
    Aug 23 at 2:17










  • I already edited, but why the answer $pi + pi k$ isn't included?
    – Mattiu
    Aug 23 at 2:18






  • 3




    The set of solutions described by your $alpha_1$ are $dots,picdot(-1),picdot 0, picdot 1, picdot 2,picdot 3,dots$. The set of solutions described by your $alpha_2$ are $dots,pi+picdot(-1),pi+picdot 0,pi+picdot 1,pi+picdot 2,dots$ which you notice after simplification is exactly the same set as the first. That doesn't make it "incorrect", just redundant.
    – JMoravitz
    Aug 23 at 2:21







  • 2




    The answer is neither $0+pi k$ nor $pi+pi k,$ but , as in the previous comment by dxiv, the answer is "$tan a=0 iff ain pi k:kin Bbb Z."$...... $pi k$ is a number. The solution is not a number. It is an infinite set of numbers.
    – DanielWainfleet
    Aug 23 at 5:46
















Check the answer again: It should be just $pi k$, since $tan$ has a period of $pi$.
– quasi
Aug 23 at 2:15





Check the answer again: It should be just $pi k$, since $tan$ has a period of $pi$.
– quasi
Aug 23 at 2:15













also it is bad form to mix degrees and radians $180 = pi$.
– Andrew Allen
Aug 23 at 2:17




also it is bad form to mix degrees and radians $180 = pi$.
– Andrew Allen
Aug 23 at 2:17












I already edited, but why the answer $pi + pi k$ isn't included?
– Mattiu
Aug 23 at 2:18




I already edited, but why the answer $pi + pi k$ isn't included?
– Mattiu
Aug 23 at 2:18




3




3




The set of solutions described by your $alpha_1$ are $dots,picdot(-1),picdot 0, picdot 1, picdot 2,picdot 3,dots$. The set of solutions described by your $alpha_2$ are $dots,pi+picdot(-1),pi+picdot 0,pi+picdot 1,pi+picdot 2,dots$ which you notice after simplification is exactly the same set as the first. That doesn't make it "incorrect", just redundant.
– JMoravitz
Aug 23 at 2:21





The set of solutions described by your $alpha_1$ are $dots,picdot(-1),picdot 0, picdot 1, picdot 2,picdot 3,dots$. The set of solutions described by your $alpha_2$ are $dots,pi+picdot(-1),pi+picdot 0,pi+picdot 1,pi+picdot 2,dots$ which you notice after simplification is exactly the same set as the first. That doesn't make it "incorrect", just redundant.
– JMoravitz
Aug 23 at 2:21





2




2




The answer is neither $0+pi k$ nor $pi+pi k,$ but , as in the previous comment by dxiv, the answer is "$tan a=0 iff ain pi k:kin Bbb Z."$...... $pi k$ is a number. The solution is not a number. It is an infinite set of numbers.
– DanielWainfleet
Aug 23 at 5:46





The answer is neither $0+pi k$ nor $pi+pi k,$ but , as in the previous comment by dxiv, the answer is "$tan a=0 iff ain pi k:kin Bbb Z."$...... $pi k$ is a number. The solution is not a number. It is an infinite set of numbers.
– DanielWainfleet
Aug 23 at 5:46











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Note that $$pi k $$ also covers $$pi + pi k $$ because $$pi + pi k = pi ( k+1) $$ and as $ k$ runs over integers, it covers $k+1$ as well.






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    Note that $$pi k $$ also covers $$pi + pi k $$ because $$pi + pi k = pi ( k+1) $$ and as $ k$ runs over integers, it covers $k+1$ as well.






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      up vote
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      down vote













      Note that $$pi k $$ also covers $$pi + pi k $$ because $$pi + pi k = pi ( k+1) $$ and as $ k$ runs over integers, it covers $k+1$ as well.






      share|cite|improve this answer






















        up vote
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        down vote










        up vote
        0
        down vote









        Note that $$pi k $$ also covers $$pi + pi k $$ because $$pi + pi k = pi ( k+1) $$ and as $ k$ runs over integers, it covers $k+1$ as well.






        share|cite|improve this answer












        Note that $$pi k $$ also covers $$pi + pi k $$ because $$pi + pi k = pi ( k+1) $$ and as $ k$ runs over integers, it covers $k+1$ as well.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 23 at 2:28









        Mohammad Riazi-Kermani

        30.1k41852




        30.1k41852



























             

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