Trig question: find all solutions

Question

2cos²4Θ + 2sin4Θ = 1

Find all solutions 0 ≤ x < 360º

If necessary, round to nearest tenth degree

 

I keep substituting cos²Θ with (1 - sin²θ).. simplifying more.. doing quadratic formula with a=2, b=-2, c=-1

 

With the answers in the back of the book I know that I'm doing something wrong :/

 

Answers should be 50.4 degrees, 84.6, 140.4, 174.6, 230.4, 264.6, 320.4, 354.6

 

I just need to know how to get these answers. :l

Answer 1

You are on the right track - if your quadratic is  

 

2sin²4x - 2sin4x - 1 = 0  (using x instead of theta because it's easier to type)  

 

then the solutions are  

 

sin4x =( 1±√3)/2  

 

(1+√3)/2 ≅ 1.4 so it is an extraneous solution since it does not have an inverse sine  

 

Now here's the tricky part:  (1 - √3)/2 ≅ -0.366 does have an inverse sine, but your calculator will only give you the principal value of the inverse sine (which is the angle between -90 and 90 degrees such that sin4x = -0.366, which is -21.47 degrees, or -21.47 + 360 = 338.53 as a positive angle).  But sine is negative in both the 3rd and 4th quadrants so there is another angle in the 3rd quadrant that also has the property that sin4x = -0.366, namely 180 + 21.47 = 201.47 degrees.  Also adding any multiple of 360 degrees to either of these 2 solutions will also be a solution.  You want solutions with x between 0 and 360 degrees so 4x will be between 0 and 1440 degrees.  We can add 360, 720 and 1080 to each of these solutions to get  

 

4x = 338.53, 698.53, 1058.53, 1418.53  and  

 

4x = 201.47, 561.47, 921.47, 1281.47  

 

Dividing by 4 gives  

 

x = 84.6, 174.6, 264.6, 354.6  and  

x = 50.4, 140.4, 230.4, 320.4