Question 1199171
.
Find the values of θ between 0° and 180° such that 2cos 3θ =3sin 3θ
~~~~~~~~~~~~~~~


<pre>
We start from this given equation

    2*cos(3θ) = 3*sin(3θ).     (1)



Looking in it, we see that cos(3θ) =/= 0  (since {{{sin^2(3theta) + cos^2(3theta)}}} = 1).



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

    {{{sin(3theta)/cos(3theta)}}} = {{{2/3}}},  or  tan(3θ) = {{{2/3}}}.    (2)


Hence,  3θ = {{{arctan(2/3)}}} = 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.



<U>ANSWER</U>.  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.
</pre>

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