SOLUTION: Find the values of θ between 0° and 180° such that 2cos 3θ =3sin 3θ

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Question 1199171: Find the values of θ between 0° and 180° such that 2cos 3θ =3sin 3θ
Found 3 solutions by math_helper, Alan3354, ikleyn:
Answer by math_helper(2461)   (Show Source): You can put this solution on YOUR website!
  

Where the graph crosses zero, the two terms are equal. The zeros highlighted are in radians, so just multiply each x value shown by to get x ( =) in degrees.

Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
Find the values of θ between 0° and 180° such that 2cos 3θ =3sin 3θ
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y = 2cos(3x) - 3sin(3x)
2*sqrt(1-sin^2(3x)) = 32in(3x)
4*(1- sin^2(3x) = 9sin^2(3x)
9sin^2(3x) + 4sin^2(3x) - 4 = 0
--------
Solve the quadratic for sin(3x)
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation (in our case ) has the following solutons:



For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=160 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: 0.480506146704084, -0.924950591148529. Here's your graph:

sin(3x) = ~0.480506
3x = 28.718
x = ~ 9.57282, 170.4272 degs
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sin(3x) = ~-0.92495
3x = -22.5536 degs = 337.446 degs
=============
Can you do the rest? And convert to radians?
I have a bad cold, and this is making me dizzy.
Answer by ikleyn(52794)   (Show Source): You can put this solution on YOUR website!
.
Find the values of θ between 0° and 180° such that 2cos 3θ =3sin 3θ
~~~~~~~~~~~~~~~ 

We start from this given equation

    2*cos(3θ) = 3*sin(3θ).     (1)



Looking in it, we see that cos(3θ) =/= 0  (since  = 1).



Therefore, we can divide both sides by cos(3θ).  Doing so, from (1) we get

     = ,  or  tan(3θ) = .    (2)


Hence,  3θ =  = 33.69 degrees is one of several possible solutions for 3θ,
which gives  θ = 33.69/3 = 11.23 degrees.



Since the tangent function is periodical with the period of 180 degrees, 
there are other solutions to equation (2)

    3θ = 33.69+180 = 213.69 degrees and  3θ = 33.69+360 = 393.69 degrees.    (3)



From (3), it gives two other solutions for θ in the interval  [0,180] degrees.


These two additional solutions are  213.69/3 = 71.23 degrees  and  393.69/3 = 131.23 degrees.



ANSWER.  In the given interval [0,180] degrees, there are three solutions to the given equation

         θ = 11.23 degrees;  θ = 71.23 degrees  and  θ = 131.23 degrees.

Solved   (in a way as it is expected and as it should be done).



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