Solving $4sin^2left(fracpi5(x-1)right)=3$ with $xin[0 , 2pi)$












0












$begingroup$



$$4sin^2left(fracpi5(x-1)right)=3qquad xin[0 , 2π)$$




My work:



$$sin^2[pi/5(x-1)] = 3/4$$



Let $p = pi/5(x-1)$



$$sin^2 p = 3/4$$



$$sin p = pm sqrt{3}/2$$



$$p = pi/3, 2pi/3, 4pi/3, 5pi/3$$



$$pi/5(x-1) = pi/3, 2pi/3, 4pi/3, 5pi/3$$



$$x = 8/3, 13/3, 23/3, 28/3$$



However, I'm not sure if these answers are correct or how many may be correct answers within the restricted range.



If, hypothetically, I wanted to set the range of $p$, what would it be without finding the correct $x$ values first?



Can you show how you would do the problem?






algebra-precalculus trigonometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    You have not found the values for x in the required range because you picked a limited set of values for p with no concern for the values of x.
    $endgroup$
    – William Elliot
    Jan 13 at 3:44












  • $begingroup$
    @WilliamElliot Could you show me how to get the full range of values for p in order to get all the x values within the restricted range?
    $endgroup$
    – andrew chen
    Jan 13 at 3:55










  • $begingroup$
    @andrewchen: $0le x<2pi$ implies $alefrac{pi}{5}(x-1)<b$ for some $a$ and $b$. Can you find what they are?
    $endgroup$
    – user587192
    Jan 13 at 4:02










  • $begingroup$
    @user587192 ohh so would it then be -π/5 ≤ p < π/5(2π-1)? In the final answer, would there then be only 2 values of x?
    $endgroup$
    – andrew chen
    Jan 13 at 4:22










  • $begingroup$
    @andrewchen: your inequality is correct. Can you find out what values of $p$ would work? And then you have the corresponding values of $x$.
    $endgroup$
    – user587192
    Jan 13 at 15:15
















0












$begingroup$



$$4sin^2left(fracpi5(x-1)right)=3qquad xin[0 , 2π)$$




My work:



$$sin^2[pi/5(x-1)] = 3/4$$



Let $p = pi/5(x-1)$



$$sin^2 p = 3/4$$



$$sin p = pm sqrt{3}/2$$



$$p = pi/3, 2pi/3, 4pi/3, 5pi/3$$



$$pi/5(x-1) = pi/3, 2pi/3, 4pi/3, 5pi/3$$



$$x = 8/3, 13/3, 23/3, 28/3$$



However, I'm not sure if these answers are correct or how many may be correct answers within the restricted range.



If, hypothetically, I wanted to set the range of $p$, what would it be without finding the correct $x$ values first?



Can you show how you would do the problem?






algebra-precalculus trigonometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    You have not found the values for x in the required range because you picked a limited set of values for p with no concern for the values of x.
    $endgroup$
    – William Elliot
    Jan 13 at 3:44












  • $begingroup$
    @WilliamElliot Could you show me how to get the full range of values for p in order to get all the x values within the restricted range?
    $endgroup$
    – andrew chen
    Jan 13 at 3:55










  • $begingroup$
    @andrewchen: $0le x<2pi$ implies $alefrac{pi}{5}(x-1)<b$ for some $a$ and $b$. Can you find what they are?
    $endgroup$
    – user587192
    Jan 13 at 4:02










  • $begingroup$
    @user587192 ohh so would it then be -π/5 ≤ p < π/5(2π-1)? In the final answer, would there then be only 2 values of x?
    $endgroup$
    – andrew chen
    Jan 13 at 4:22










  • $begingroup$
    @andrewchen: your inequality is correct. Can you find out what values of $p$ would work? And then you have the corresponding values of $x$.
    $endgroup$
    – user587192
    Jan 13 at 15:15














0












0








0





$begingroup$



$$4sin^2left(fracpi5(x-1)right)=3qquad xin[0 , 2π)$$




My work:



$$sin^2[pi/5(x-1)] = 3/4$$



Let $p = pi/5(x-1)$



$$sin^2 p = 3/4$$



$$sin p = pm sqrt{3}/2$$



$$p = pi/3, 2pi/3, 4pi/3, 5pi/3$$



$$pi/5(x-1) = pi/3, 2pi/3, 4pi/3, 5pi/3$$



$$x = 8/3, 13/3, 23/3, 28/3$$



However, I'm not sure if these answers are correct or how many may be correct answers within the restricted range.



If, hypothetically, I wanted to set the range of $p$, what would it be without finding the correct $x$ values first?



Can you show how you would do the problem?






algebra-precalculus trigonometry







share|cite|improve this question











$endgroup$





$$4sin^2left(fracpi5(x-1)right)=3qquad xin[0 , 2π)$$




My work:



$$sin^2[pi/5(x-1)] = 3/4$$



Let $p = pi/5(x-1)$



$$sin^2 p = 3/4$$



$$sin p = pm sqrt{3}/2$$



$$p = pi/3, 2pi/3, 4pi/3, 5pi/3$$



$$pi/5(x-1) = pi/3, 2pi/3, 4pi/3, 5pi/3$$



$$x = 8/3, 13/3, 23/3, 28/3$$



However, I'm not sure if these answers are correct or how many may be correct answers within the restricted range.



If, hypothetically, I wanted to set the range of $p$, what would it be without finding the correct $x$ values first?



Can you show how you would do the problem?






algebra-precalculus trigonometry




algebra-precalculus trigonometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 13 at 3:51







andrew chen

















asked Jan 13 at 3:14









andrew chenandrew chen

114




114












  • $begingroup$
    You have not found the values for x in the required range because you picked a limited set of values for p with no concern for the values of x.
    $endgroup$
    – William Elliot
    Jan 13 at 3:44












  • $begingroup$
    @WilliamElliot Could you show me how to get the full range of values for p in order to get all the x values within the restricted range?
    $endgroup$
    – andrew chen
    Jan 13 at 3:55










  • $begingroup$
    @andrewchen: $0le x<2pi$ implies $alefrac{pi}{5}(x-1)<b$ for some $a$ and $b$. Can you find what they are?
    $endgroup$
    – user587192
    Jan 13 at 4:02










  • $begingroup$
    @user587192 ohh so would it then be -π/5 ≤ p < π/5(2π-1)? In the final answer, would there then be only 2 values of x?
    $endgroup$
    – andrew chen
    Jan 13 at 4:22










  • $begingroup$
    @andrewchen: your inequality is correct. Can you find out what values of $p$ would work? And then you have the corresponding values of $x$.
    $endgroup$
    – user587192
    Jan 13 at 15:15


















  • $begingroup$
    You have not found the values for x in the required range because you picked a limited set of values for p with no concern for the values of x.
    $endgroup$
    – William Elliot
    Jan 13 at 3:44












  • $begingroup$
    @WilliamElliot Could you show me how to get the full range of values for p in order to get all the x values within the restricted range?
    $endgroup$
    – andrew chen
    Jan 13 at 3:55










  • $begingroup$
    @andrewchen: $0le x<2pi$ implies $alefrac{pi}{5}(x-1)<b$ for some $a$ and $b$. Can you find what they are?
    $endgroup$
    – user587192
    Jan 13 at 4:02










  • $begingroup$
    @user587192 ohh so would it then be -π/5 ≤ p < π/5(2π-1)? In the final answer, would there then be only 2 values of x?
    $endgroup$
    – andrew chen
    Jan 13 at 4:22










  • $begingroup$
    @andrewchen: your inequality is correct. Can you find out what values of $p$ would work? And then you have the corresponding values of $x$.
    $endgroup$
    – user587192
    Jan 13 at 15:15
















$begingroup$
You have not found the values for x in the required range because you picked a limited set of values for p with no concern for the values of x.
$endgroup$
– William Elliot
Jan 13 at 3:44






$begingroup$
You have not found the values for x in the required range because you picked a limited set of values for p with no concern for the values of x.
$endgroup$
– William Elliot
Jan 13 at 3:44














$begingroup$
@WilliamElliot Could you show me how to get the full range of values for p in order to get all the x values within the restricted range?
$endgroup$
– andrew chen
Jan 13 at 3:55




$begingroup$
@WilliamElliot Could you show me how to get the full range of values for p in order to get all the x values within the restricted range?
$endgroup$
– andrew chen
Jan 13 at 3:55












$begingroup$
@andrewchen: $0le x<2pi$ implies $alefrac{pi}{5}(x-1)<b$ for some $a$ and $b$. Can you find what they are?
$endgroup$
– user587192
Jan 13 at 4:02




$begingroup$
@andrewchen: $0le x<2pi$ implies $alefrac{pi}{5}(x-1)<b$ for some $a$ and $b$. Can you find what they are?
$endgroup$
– user587192
Jan 13 at 4:02












$begingroup$
@user587192 ohh so would it then be -π/5 ≤ p < π/5(2π-1)? In the final answer, would there then be only 2 values of x?
$endgroup$
– andrew chen
Jan 13 at 4:22




$begingroup$
@user587192 ohh so would it then be -π/5 ≤ p < π/5(2π-1)? In the final answer, would there then be only 2 values of x?
$endgroup$
– andrew chen
Jan 13 at 4:22












$begingroup$
@andrewchen: your inequality is correct. Can you find out what values of $p$ would work? And then you have the corresponding values of $x$.
$endgroup$
– user587192
Jan 13 at 15:15




$begingroup$
@andrewchen: your inequality is correct. Can you find out what values of $p$ would work? And then you have the corresponding values of $x$.
$endgroup$
– user587192
Jan 13 at 15:15










3 Answers
3






active

oldest

votes


















1












$begingroup$

Since $xin[0,2pi)$, you must have
$$
p=frac{pi}{5}(x-1)in[a,b)
$$
for some $a$ and $b$. Find $a$ and $b$ and then find $p$ such that $pin[a,b)$ and
$$
sin p=pmfrac{sqrt{3}}{2}.
$$
Once you have the values of $p$ in $[a,b)$, use the relation
$$
p=frac{pi}{5}(x-1)
$$
to the values of $x$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    For $sin^2y=sin^2Aiffcos^2y=cos^2Aifftan^2y=tan^2A$



    Method$#1:$ use Prove $ sin(A+B)sin(A-B)=sin^2A-sin^2B $



    $sin(ypm A)=0,y=mpimp A$



    Method$#2:$



    Use $cos2x=1-2sin^2x,$



    and $cos2x=cos2Aimplies2x=2npipm2A$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Here $$A=dfracpi3$$
      $endgroup$
      – lab bhattacharjee
      Jan 13 at 5:23



















    0












    $begingroup$

    When you get to $sin^2left({piover5}{(x-1)}right)={3over4}$,so using $sin^2(x)+cos^2(x)=1$,$cosleft({piover5}{(x-1)}right)={1over2}$,$left({piover5}{(x-1)}right)={piover3}+2pi n$. It is easy from here now.






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Since $xin[0,2pi)$, you must have
      $$
      p=frac{pi}{5}(x-1)in[a,b)
      $$
      for some $a$ and $b$. Find $a$ and $b$ and then find $p$ such that $pin[a,b)$ and
      $$
      sin p=pmfrac{sqrt{3}}{2}.
      $$
      Once you have the values of $p$ in $[a,b)$, use the relation
      $$
      p=frac{pi}{5}(x-1)
      $$
      to the values of $x$.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Since $xin[0,2pi)$, you must have
        $$
        p=frac{pi}{5}(x-1)in[a,b)
        $$
        for some $a$ and $b$. Find $a$ and $b$ and then find $p$ such that $pin[a,b)$ and
        $$
        sin p=pmfrac{sqrt{3}}{2}.
        $$
        Once you have the values of $p$ in $[a,b)$, use the relation
        $$
        p=frac{pi}{5}(x-1)
        $$
        to the values of $x$.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Since $xin[0,2pi)$, you must have
          $$
          p=frac{pi}{5}(x-1)in[a,b)
          $$
          for some $a$ and $b$. Find $a$ and $b$ and then find $p$ such that $pin[a,b)$ and
          $$
          sin p=pmfrac{sqrt{3}}{2}.
          $$
          Once you have the values of $p$ in $[a,b)$, use the relation
          $$
          p=frac{pi}{5}(x-1)
          $$
          to the values of $x$.






          share|cite|improve this answer









          $endgroup$



          Since $xin[0,2pi)$, you must have
          $$
          p=frac{pi}{5}(x-1)in[a,b)
          $$
          for some $a$ and $b$. Find $a$ and $b$ and then find $p$ such that $pin[a,b)$ and
          $$
          sin p=pmfrac{sqrt{3}}{2}.
          $$
          Once you have the values of $p$ in $[a,b)$, use the relation
          $$
          p=frac{pi}{5}(x-1)
          $$
          to the values of $x$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 13 at 3:47









          user587192user587192

          1,868315




          1,868315























              0












              $begingroup$

              For $sin^2y=sin^2Aiffcos^2y=cos^2Aifftan^2y=tan^2A$



              Method$#1:$ use Prove $ sin(A+B)sin(A-B)=sin^2A-sin^2B $



              $sin(ypm A)=0,y=mpimp A$



              Method$#2:$



              Use $cos2x=1-2sin^2x,$



              and $cos2x=cos2Aimplies2x=2npipm2A$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Here $$A=dfracpi3$$
                $endgroup$
                – lab bhattacharjee
                Jan 13 at 5:23
















              0












              $begingroup$

              For $sin^2y=sin^2Aiffcos^2y=cos^2Aifftan^2y=tan^2A$



              Method$#1:$ use Prove $ sin(A+B)sin(A-B)=sin^2A-sin^2B $



              $sin(ypm A)=0,y=mpimp A$



              Method$#2:$



              Use $cos2x=1-2sin^2x,$



              and $cos2x=cos2Aimplies2x=2npipm2A$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Here $$A=dfracpi3$$
                $endgroup$
                – lab bhattacharjee
                Jan 13 at 5:23














              0












              0








              0





              $begingroup$

              For $sin^2y=sin^2Aiffcos^2y=cos^2Aifftan^2y=tan^2A$



              Method$#1:$ use Prove $ sin(A+B)sin(A-B)=sin^2A-sin^2B $



              $sin(ypm A)=0,y=mpimp A$



              Method$#2:$



              Use $cos2x=1-2sin^2x,$



              and $cos2x=cos2Aimplies2x=2npipm2A$






              share|cite|improve this answer









              $endgroup$



              For $sin^2y=sin^2Aiffcos^2y=cos^2Aifftan^2y=tan^2A$



              Method$#1:$ use Prove $ sin(A+B)sin(A-B)=sin^2A-sin^2B $



              $sin(ypm A)=0,y=mpimp A$



              Method$#2:$



              Use $cos2x=1-2sin^2x,$



              and $cos2x=cos2Aimplies2x=2npipm2A$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 13 at 4:18









              lab bhattacharjeelab bhattacharjee

              225k15157275




              225k15157275












              • $begingroup$
                Here $$A=dfracpi3$$
                $endgroup$
                – lab bhattacharjee
                Jan 13 at 5:23


















              • $begingroup$
                Here $$A=dfracpi3$$
                $endgroup$
                – lab bhattacharjee
                Jan 13 at 5:23
















              $begingroup$
              Here $$A=dfracpi3$$
              $endgroup$
              – lab bhattacharjee
              Jan 13 at 5:23




              $begingroup$
              Here $$A=dfracpi3$$
              $endgroup$
              – lab bhattacharjee
              Jan 13 at 5:23











              0












              $begingroup$

              When you get to $sin^2left({piover5}{(x-1)}right)={3over4}$,so using $sin^2(x)+cos^2(x)=1$,$cosleft({piover5}{(x-1)}right)={1over2}$,$left({piover5}{(x-1)}right)={piover3}+2pi n$. It is easy from here now.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                When you get to $sin^2left({piover5}{(x-1)}right)={3over4}$,so using $sin^2(x)+cos^2(x)=1$,$cosleft({piover5}{(x-1)}right)={1over2}$,$left({piover5}{(x-1)}right)={piover3}+2pi n$. It is easy from here now.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  When you get to $sin^2left({piover5}{(x-1)}right)={3over4}$,so using $sin^2(x)+cos^2(x)=1$,$cosleft({piover5}{(x-1)}right)={1over2}$,$left({piover5}{(x-1)}right)={piover3}+2pi n$. It is easy from here now.






                  share|cite|improve this answer









                  $endgroup$



                  When you get to $sin^2left({piover5}{(x-1)}right)={3over4}$,so using $sin^2(x)+cos^2(x)=1$,$cosleft({piover5}{(x-1)}right)={1over2}$,$left({piover5}{(x-1)}right)={piover3}+2pi n$. It is easy from here now.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 14 at 2:14









                  Math LoverMath Lover

                  877




                  877






























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