I have another quadratic trig equation issue.

Question

-4cos^2x+4sqrt(2)sin (x) +6=0

Pi/2 <x <3pi/2

 

I don't understand what I should do with the square root.

Answer 1

-4cos²(x) + 4√(2)sin(x) + 6 = 0

 

We want to have the quadratic in terms o sine or cosine.  We will get everything in terms of sine.

 

We can use the identity  sin²(x) + cos²(x) =

 

 

-4(1 - sin²(x)) + 4√(2)sin(x) + 6 = 0

 

-4 + 4sin²(x) + 4√(2)sin(x) + 6 = 0

 

 

4sin²(x) + 4√(2)sin(x) + 2 = 0

 

 

Factor out the 2.  Let  x = sin(x).

 

2(2x² + 2√(2)x + 1) = 0

 

 

2x² + 2√(2)x + 1 = 0

 

Use the quadratic formula to solve for x.

 

x = (-8 ± √(8 - 8)) / 4

x = -32 / 4

x = -8

 

sin(x) = -8               and         sin²(x) = (-8)²

                                             sin²(x) = 64

                                       

 

There is no angle that when it is sined, will result in negative -8.  Therefore, there are no solutions to this problem.