Solve on the interval $[0,2pi)$: $4 sin(x) cos(x)=1$. [closed]












-1














I tried using the product-to-sum formulas, but did not come up with the correct answer.










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closed as off-topic by Shailesh, amWhy, Toby Mak, Delta-u, José Carlos Santos Nov 29 '18 at 12:34


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


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    -1














    I tried using the product-to-sum formulas, but did not come up with the correct answer.










    share|cite|improve this question















    closed as off-topic by Shailesh, amWhy, Toby Mak, Delta-u, José Carlos Santos Nov 29 '18 at 12:34


    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." – Shailesh, amWhy, Toby Mak, Delta-u, José Carlos Santos

    If this question can be reworded to fit the rules in the help center, please edit the question.
















      -1












      -1








      -1







      I tried using the product-to-sum formulas, but did not come up with the correct answer.










      share|cite|improve this question















      I tried using the product-to-sum formulas, but did not come up with the correct answer.







      algebra-precalculus trigonometry






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      edited Nov 29 '18 at 10:52









      amWhy

      192k28225439




      192k28225439










      asked Nov 29 '18 at 0:39









      math818math818

      33




      33




      closed as off-topic by Shailesh, amWhy, Toby Mak, Delta-u, José Carlos Santos Nov 29 '18 at 12:34


      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." – Shailesh, amWhy, Toby Mak, Delta-u, José Carlos Santos

      If this question can be reworded to fit the rules in the help center, please edit the question.




      closed as off-topic by Shailesh, amWhy, Toby Mak, Delta-u, José Carlos Santos Nov 29 '18 at 12:34


      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." – Shailesh, amWhy, Toby Mak, Delta-u, José Carlos Santos

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          2 Answers
          2






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          1














          As was hinted, $$sin2x=2sin xcos x$$
          Hence, your equation becomes
          $$2sin2x=1$$
          $$sin2x=frac12$$
          $$2x=arcsinfrac12$$
          $$2x=fracpi6,,frac{5pi}6$$
          $$x=fracpi{12},,frac{5pi}{12}$$






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            5














            Hint: use the double-angle formula for sine:



            $$sin(2theta) = 2sin(theta)cos(theta)$$






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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              1














              As was hinted, $$sin2x=2sin xcos x$$
              Hence, your equation becomes
              $$2sin2x=1$$
              $$sin2x=frac12$$
              $$2x=arcsinfrac12$$
              $$2x=fracpi6,,frac{5pi}6$$
              $$x=fracpi{12},,frac{5pi}{12}$$






              share|cite|improve this answer




























                1














                As was hinted, $$sin2x=2sin xcos x$$
                Hence, your equation becomes
                $$2sin2x=1$$
                $$sin2x=frac12$$
                $$2x=arcsinfrac12$$
                $$2x=fracpi6,,frac{5pi}6$$
                $$x=fracpi{12},,frac{5pi}{12}$$






                share|cite|improve this answer


























                  1












                  1








                  1






                  As was hinted, $$sin2x=2sin xcos x$$
                  Hence, your equation becomes
                  $$2sin2x=1$$
                  $$sin2x=frac12$$
                  $$2x=arcsinfrac12$$
                  $$2x=fracpi6,,frac{5pi}6$$
                  $$x=fracpi{12},,frac{5pi}{12}$$






                  share|cite|improve this answer














                  As was hinted, $$sin2x=2sin xcos x$$
                  Hence, your equation becomes
                  $$2sin2x=1$$
                  $$sin2x=frac12$$
                  $$2x=arcsinfrac12$$
                  $$2x=fracpi6,,frac{5pi}6$$
                  $$x=fracpi{12},,frac{5pi}{12}$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 29 '18 at 1:55

























                  answered Nov 29 '18 at 1:36









                  clathratusclathratus

                  3,325331




                  3,325331























                      5














                      Hint: use the double-angle formula for sine:



                      $$sin(2theta) = 2sin(theta)cos(theta)$$






                      share|cite|improve this answer




























                        5














                        Hint: use the double-angle formula for sine:



                        $$sin(2theta) = 2sin(theta)cos(theta)$$






                        share|cite|improve this answer


























                          5












                          5








                          5






                          Hint: use the double-angle formula for sine:



                          $$sin(2theta) = 2sin(theta)cos(theta)$$






                          share|cite|improve this answer














                          Hint: use the double-angle formula for sine:



                          $$sin(2theta) = 2sin(theta)cos(theta)$$







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Nov 29 '18 at 0:42









                          Bernard

                          118k639112




                          118k639112










                          answered Nov 29 '18 at 0:40









                          Eevee TrainerEevee Trainer

                          5,0471734




                          5,0471734















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