SOLUTION: Determine the values of 2θ (not θ) on [0,2π) that satisfy the following equation. (Separate multiple solutions with a comma. Give exact answers.) 3sin(2θ)=&

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Question 1101623: Determine the values of
2θ (not θ) on [0,2π) that satisfy the following equation. (Separate multiple solutions with a comma. Give exact answers.)
3sin(2θ)=−3/√2
2θ=
You now have two equations representing all possible solutions for 2θ. Solve each of those equations for θ. (Let θ1 and θ2 represent the solutions on [0,2π), where θ1is less than θ2.)
θ1=

θ2=
Use these general solutions for θ to find the four solutions to 3sin(2θ)=−3/√2 on the intervall
[0,2π). (Separate multiple solutions with a comma. Give exact answers.)
θ=

Found 2 solutions by stanbon, ikleyn:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
Determine the values of 2t (not t) on [0,2π) that satisfy the following equation. (Separate multiple solutions with a comma. Give exact answers.)
3sin(2t)=−3/√2
sin(2t) = -1/sqrt(2)
2t = 5pi/4 or 2t = 7pi/4
-----------------------------------------------------

You now have two equations representing all possible solutions for 2θ. Solve each of those equations for θ. (Let θ1 and θ2 represent the solutions on [0,2π), where θ1is less than θ2.)
t1= 5pi/8
t2= 7pi/8
--------------------------------
Cheers,
Stan H.
--------------
Use these general solutions for θ to find the four solutions to 3sin(2θ)=−3/√2 on the interval
[0,2π). (Separate multiple solutions with a comma. Give exact answers.)
θ=

Answer by ikleyn(52832)   (Show Source): You can put this solution on YOUR website!
.
3*sin(2*a) = -3/sqrt(2)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~


        Due to "typography" issues,  I will replace    in my post by simple  "a".


 =       (1)       ====>  (divide both sides by 3)  ====>


 = ,   or, which is the same,

 = .


It implies   =    or    = .



    Everything was simple to this point. 

    But in reality, accurate analysis only  STARTS  from this point.


1)  It is obvious that   =   implies   = . 

    But if you stop here, you will loose another existing solution of the same family.

    It is   =  = .

    Indeed,   =  =  is GEOMETRICALLY the same angle as   and has the same value of sine,

    so  is the solution to the original equation  (1), too.


    Thus the relation   =   creates and generates not one solution , but TWO solutions    and    

    of the same family.     Notice, that they BOTH belong to the interval  [0,).



2)  The same or the similar story is with the solution   = .


    It is obvious that   =   implies   = . 

    But if you stop here, you will loose another existing solution of the same family.

    It is   =  = .

    Indeed,   =  =   is GEOMETRICALLY the same angle as   and has the same value of sine,

    so  is the solution to the original equation  (1), too.


    Thus the relation   =   creates and generates not one solution , but TWO solutions    and    

    of the same family.     Notice, that they BOTH belong to the interval  [0,).



3.  Thus the original equation (1) has 4 (four, FOUR) solutions in the interval  [0,):

    ,  ,    and  .



4.  The plot below visually confirms existing of 4 solutions to the given equality:





Plot y =   (red)  and y =  (green)


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