Question 1117559
Could someone give me some tips about how to solve this question ? I don't know how to do this question with decimals. Thanks you!
Give the smallest two solutions of cos(6θ) = 0.2771 on [ 0,2π ).
Separate the two solutions with a comma. 
Be sure to round only once at the end.

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The last sentence needs to specify how many decimal places (which I'd expect to be 4 or less, because the value 0.2771 has 4 signficant digits).  I will carry 7 decimal digits for all the calculations and you can round to the required number of digits.  <br>

Step-by-step:
{{{ cos(6*theta) = 0.2771 }}}
{{{ cos^-1(cos(6*theta)) = cos^-1(0.2771) }}}
{{{ 6*theta = 1.2900217 }}} <br>

Now, there are two solutions, the one above (in Q1), and {{{6*theta = 2pi - 1.2900217 }}}  (in Q4). <br>

This is because x is positive in Q1 and Q4, thus {{{ cos(theta) = x/r }}} has  positive x in those two quadrants.  I am referring to {{ 6* theta }}} here, finding {{{ theta }}} itself requires dividing by 6.

Solving the two cases for {{{theta}}} you should get  {{{ highlight( theta = 0.2150036 )}}} rad  for the Q1 angle and {{{ highlight( theta =  0.8321938 )}}} rad for the Q4 angle.<br>

A graph might help visualize this, where A and B are the approximate locations of the two solutions:<br>

{{{ drawing(500,300, -0.1, 2, -2, 2, grid(1),
graph(500, 300 ,  -0.1, 2, -2, 2, grid(0),
         cos(6*x) ),
    locate(0.2150, 0.3771, "A"),
    locate(0.8322, 0.3771, "B")
)
}}}

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EDIT 5/30/18:
In my post, I was referring to the quadrants of {{{ 6*theta }}} not {{{ theta }}} itself.  For clarity, let {{{ alpha = 6*theta}}} then it was {{{ alpha }}} in Q1 and Q4.   I did mention this fact and I apologize if it was not clear.