How to find $x$ such that $cos 2x - cos4x = frac12$ and $0<x<3$? [closed]

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How to solve this trigonometric equation, giving your answer in exact form:



$$cos 2x - cos4x = frac12$$ where $0 < x <3$










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closed as off-topic by amWhy, José Carlos Santos, user99914, Deepesh Meena, user91500 Sep 10 at 9:27


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, Community, Deepesh Meena, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.












  • You could try using the double angle formula for cos in terms of cos and solve it as a quadratic equation.
    – Prashin Jeevaganth
    Sep 9 at 13:58










  • Use the half-angle formula for $cos(4x)$ to get a quadratic equation in $cos(2x)$
    – Claude Leibovici
    Sep 9 at 13:58






  • 1




    We're not doing your homework for you without you showing us your efforts.
    – TheSimpliFire
    Sep 9 at 14:00














up vote
-1
down vote

favorite












How to solve this trigonometric equation, giving your answer in exact form:



$$cos 2x - cos4x = frac12$$ where $0 < x <3$










share|cite|improve this question















closed as off-topic by amWhy, José Carlos Santos, user99914, Deepesh Meena, user91500 Sep 10 at 9:27


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, Community, Deepesh Meena, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.












  • You could try using the double angle formula for cos in terms of cos and solve it as a quadratic equation.
    – Prashin Jeevaganth
    Sep 9 at 13:58










  • Use the half-angle formula for $cos(4x)$ to get a quadratic equation in $cos(2x)$
    – Claude Leibovici
    Sep 9 at 13:58






  • 1




    We're not doing your homework for you without you showing us your efforts.
    – TheSimpliFire
    Sep 9 at 14:00












up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











How to solve this trigonometric equation, giving your answer in exact form:



$$cos 2x - cos4x = frac12$$ where $0 < x <3$










share|cite|improve this question















How to solve this trigonometric equation, giving your answer in exact form:



$$cos 2x - cos4x = frac12$$ where $0 < x <3$







trigonometry






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share|cite|improve this question













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edited Sep 9 at 14:41









Bernard

112k636105




112k636105










asked Sep 9 at 13:55









Andrew Hoang

4




4




closed as off-topic by amWhy, José Carlos Santos, user99914, Deepesh Meena, user91500 Sep 10 at 9:27


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, Community, Deepesh Meena, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by amWhy, José Carlos Santos, user99914, Deepesh Meena, user91500 Sep 10 at 9:27


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, Community, Deepesh Meena, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.











  • You could try using the double angle formula for cos in terms of cos and solve it as a quadratic equation.
    – Prashin Jeevaganth
    Sep 9 at 13:58










  • Use the half-angle formula for $cos(4x)$ to get a quadratic equation in $cos(2x)$
    – Claude Leibovici
    Sep 9 at 13:58






  • 1




    We're not doing your homework for you without you showing us your efforts.
    – TheSimpliFire
    Sep 9 at 14:00
















  • You could try using the double angle formula for cos in terms of cos and solve it as a quadratic equation.
    – Prashin Jeevaganth
    Sep 9 at 13:58










  • Use the half-angle formula for $cos(4x)$ to get a quadratic equation in $cos(2x)$
    – Claude Leibovici
    Sep 9 at 13:58






  • 1




    We're not doing your homework for you without you showing us your efforts.
    – TheSimpliFire
    Sep 9 at 14:00















You could try using the double angle formula for cos in terms of cos and solve it as a quadratic equation.
– Prashin Jeevaganth
Sep 9 at 13:58




You could try using the double angle formula for cos in terms of cos and solve it as a quadratic equation.
– Prashin Jeevaganth
Sep 9 at 13:58












Use the half-angle formula for $cos(4x)$ to get a quadratic equation in $cos(2x)$
– Claude Leibovici
Sep 9 at 13:58




Use the half-angle formula for $cos(4x)$ to get a quadratic equation in $cos(2x)$
– Claude Leibovici
Sep 9 at 13:58




1




1




We're not doing your homework for you without you showing us your efforts.
– TheSimpliFire
Sep 9 at 14:00




We're not doing your homework for you without you showing us your efforts.
– TheSimpliFire
Sep 9 at 14:00










4 Answers
4






active

oldest

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up vote
1
down vote













Just for the first step.



beginalign
cos2x-cos4x&=frac12 \
cos2x-(cos^22x-1)&=frac12 \
endalign






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    up vote
    0
    down vote













    HINT



    Recall that



    $$cos 4x=2cos^2 2x-1$$



    then use quadratic equation.






    share|cite|improve this answer



























      up vote
      0
      down vote













      Use that $$cos(x)-cos(y)=-2 sin left(fracx2-fracy2right) sin
      left(fracx2+fracy2right)$$






      share|cite|improve this answer



























        up vote
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        Notice that $$cos 4x=2cos^2 2x-1,~~~forall x in mathbbR,$$



        Hence, the equation could be rewritten as $$cos 2x-(2cos^2 2x-1)=frac12,$$
        namely $$4(cos 2x)^2-2cos 2x-1=0.$$
        Therefore $$cos 2x=frac1pm sqrt54,$$
        which implies $$2x=arccosfrac1pm sqrt54,$$ or
        $$2x=2pi-arccosfrac1pm sqrt54.$$
        Notice that $0<x<3$. As a result, $$x=frac12arccosfrac1pm sqrt54,$$or $$x=pi-frac12arccosfrac1pm sqrt54.$$






        share|cite|improve this answer





























          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          1
          down vote













          Just for the first step.



          beginalign
          cos2x-cos4x&=frac12 \
          cos2x-(cos^22x-1)&=frac12 \
          endalign






          share|cite|improve this answer
























            up vote
            1
            down vote













            Just for the first step.



            beginalign
            cos2x-cos4x&=frac12 \
            cos2x-(cos^22x-1)&=frac12 \
            endalign






            share|cite|improve this answer






















              up vote
              1
              down vote










              up vote
              1
              down vote









              Just for the first step.



              beginalign
              cos2x-cos4x&=frac12 \
              cos2x-(cos^22x-1)&=frac12 \
              endalign






              share|cite|improve this answer












              Just for the first step.



              beginalign
              cos2x-cos4x&=frac12 \
              cos2x-(cos^22x-1)&=frac12 \
              endalign







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Sep 9 at 13:59









              Yuta

              63929




              63929




















                  up vote
                  0
                  down vote













                  HINT



                  Recall that



                  $$cos 4x=2cos^2 2x-1$$



                  then use quadratic equation.






                  share|cite|improve this answer
























                    up vote
                    0
                    down vote













                    HINT



                    Recall that



                    $$cos 4x=2cos^2 2x-1$$



                    then use quadratic equation.






                    share|cite|improve this answer






















                      up vote
                      0
                      down vote










                      up vote
                      0
                      down vote









                      HINT



                      Recall that



                      $$cos 4x=2cos^2 2x-1$$



                      then use quadratic equation.






                      share|cite|improve this answer












                      HINT



                      Recall that



                      $$cos 4x=2cos^2 2x-1$$



                      then use quadratic equation.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Sep 9 at 14:00









                      gimusi

                      74.4k73889




                      74.4k73889




















                          up vote
                          0
                          down vote













                          Use that $$cos(x)-cos(y)=-2 sin left(fracx2-fracy2right) sin
                          left(fracx2+fracy2right)$$






                          share|cite|improve this answer
























                            up vote
                            0
                            down vote













                            Use that $$cos(x)-cos(y)=-2 sin left(fracx2-fracy2right) sin
                            left(fracx2+fracy2right)$$






                            share|cite|improve this answer






















                              up vote
                              0
                              down vote










                              up vote
                              0
                              down vote









                              Use that $$cos(x)-cos(y)=-2 sin left(fracx2-fracy2right) sin
                              left(fracx2+fracy2right)$$






                              share|cite|improve this answer












                              Use that $$cos(x)-cos(y)=-2 sin left(fracx2-fracy2right) sin
                              left(fracx2+fracy2right)$$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Sep 9 at 14:01









                              Dr. Sonnhard Graubner

                              69.1k32761




                              69.1k32761




















                                  up vote
                                  0
                                  down vote













                                  Notice that $$cos 4x=2cos^2 2x-1,~~~forall x in mathbbR,$$



                                  Hence, the equation could be rewritten as $$cos 2x-(2cos^2 2x-1)=frac12,$$
                                  namely $$4(cos 2x)^2-2cos 2x-1=0.$$
                                  Therefore $$cos 2x=frac1pm sqrt54,$$
                                  which implies $$2x=arccosfrac1pm sqrt54,$$ or
                                  $$2x=2pi-arccosfrac1pm sqrt54.$$
                                  Notice that $0<x<3$. As a result, $$x=frac12arccosfrac1pm sqrt54,$$or $$x=pi-frac12arccosfrac1pm sqrt54.$$






                                  share|cite|improve this answer


























                                    up vote
                                    0
                                    down vote













                                    Notice that $$cos 4x=2cos^2 2x-1,~~~forall x in mathbbR,$$



                                    Hence, the equation could be rewritten as $$cos 2x-(2cos^2 2x-1)=frac12,$$
                                    namely $$4(cos 2x)^2-2cos 2x-1=0.$$
                                    Therefore $$cos 2x=frac1pm sqrt54,$$
                                    which implies $$2x=arccosfrac1pm sqrt54,$$ or
                                    $$2x=2pi-arccosfrac1pm sqrt54.$$
                                    Notice that $0<x<3$. As a result, $$x=frac12arccosfrac1pm sqrt54,$$or $$x=pi-frac12arccosfrac1pm sqrt54.$$






                                    share|cite|improve this answer
























                                      up vote
                                      0
                                      down vote










                                      up vote
                                      0
                                      down vote









                                      Notice that $$cos 4x=2cos^2 2x-1,~~~forall x in mathbbR,$$



                                      Hence, the equation could be rewritten as $$cos 2x-(2cos^2 2x-1)=frac12,$$
                                      namely $$4(cos 2x)^2-2cos 2x-1=0.$$
                                      Therefore $$cos 2x=frac1pm sqrt54,$$
                                      which implies $$2x=arccosfrac1pm sqrt54,$$ or
                                      $$2x=2pi-arccosfrac1pm sqrt54.$$
                                      Notice that $0<x<3$. As a result, $$x=frac12arccosfrac1pm sqrt54,$$or $$x=pi-frac12arccosfrac1pm sqrt54.$$






                                      share|cite|improve this answer














                                      Notice that $$cos 4x=2cos^2 2x-1,~~~forall x in mathbbR,$$



                                      Hence, the equation could be rewritten as $$cos 2x-(2cos^2 2x-1)=frac12,$$
                                      namely $$4(cos 2x)^2-2cos 2x-1=0.$$
                                      Therefore $$cos 2x=frac1pm sqrt54,$$
                                      which implies $$2x=arccosfrac1pm sqrt54,$$ or
                                      $$2x=2pi-arccosfrac1pm sqrt54.$$
                                      Notice that $0<x<3$. As a result, $$x=frac12arccosfrac1pm sqrt54,$$or $$x=pi-frac12arccosfrac1pm sqrt54.$$







                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited Sep 9 at 14:50

























                                      answered Sep 9 at 14:45









                                      mengdie1982

                                      3,774216




                                      3,774216












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