Thread: Solving Quadratic Trig Equations Algebraically

Solve the following quadratic trig equations algebraically (without graphing technology) for the domain 0 < x < 360°. Use exact solutions whenever possible.

6cos^2(x) + cos(x) - 1 = 0


I have done the following:

6cos^2(x) + cos(x) - 1 = 0
6cos^2(x) + 3cos(x) - 2cos(x) - 1 = 0
3cos(x) [2cos(x) + 1] -1 [2cos(x) +1] = 0
[3cos(x) -1] [2cos(x) + 1] = 0

3cos(x) - 1 = 0
3cos(x) = 1
cos(x) = 1/3

Quadrants I and IV
Reference Angle: ???

At this point I think I've done something wrong because there is no reference angle for cos(x) = 1/3.


2cos(x) + 1 = 0
2cos(x) = -1
cos(x) = -1/2

Quadrants II and III
Reference Angle= 60 degrees

180 - 60 = 120 degrees
180 + 60 = 240 degrees



Therefore, so far, I have solved for the values: 120 degrees and 240 degrees

Originally Posted by starshine84

Solve the following quadratic trig equations algebraically (without graphing technology) for the domain 0 < x < 360°. Use exact solutions whenever possible.

6cos^2(x) + cos(x) - 1 = 0


I have done the following:

6cos^2(x) + cos(x) - 1 = 0
6cos^2(x) + 3cos(x) - 2cos(x) - 1 = 0
3cos(x) [2cos(x) + 1] -1 [2cos(x) +1] = 0
[3cos(x) -1] [2cos(x) + 1] = 0

3cos(x) - 1 = 0
3cos(x) = 1
cos(x) = 1/3

Quadrants I and IV
Reference Angle: ???

At this point I think I've done something wrong because there is no reference angle for cos(x) = 1/3.

Why not? 1/3 is between -1 and 1, of course, there is a reference angle. Any calculator will tell you that cos(70.53)= 1/3 (approximately). The problem did say "use exact solutions whenever possible.
Or, if you prefer, exact solutions would be and .

2cos(x) + 1 = 0
2cos(x) = -1
cos(x) = -1/2

Quadrants II and III
Reference Angle= 60 degrees

180 - 60 = 120 degrees
180 + 60 = 240 degrees



Therefore, so far, I have solved for the values: 120 degrees and 240 degrees

So are my final solutions x= 70.53, 120, 240, and 289.47??

Thank you very much for you help!