SOLUTION: Find, to the nearest degree, all values of x in the interval 0° ≤ x ≤ 360° that satisfy the equation 3 cos 2x + cos x + 2 = 0. Please check my answers to see if I a

Algebra ->  Trigonometry-basics -> SOLUTION: Find, to the nearest degree, all values of x in the interval 0° ≤ x ≤ 360° that satisfy the equation 3 cos 2x + cos x + 2 = 0. Please check my answers to see if I a      Log On


   

Question 389120: Find, to the nearest degree, all values of x in the interval 0° ≤ x ≤ 360° that satisfy the equation 3 cos 2x + cos x + 2 = 0.
Please check my answers to see if I am correct: 71°; 60°; 300° and 289°
Thanks
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
You got two right and two wrong!

3Cos%282x%29+%2B+Cos%28x%29+%2B+2+=+0

Use the identity Cos%282x%29=2Cos%5E2x-1 

3%282Cos%5E2x-1%29+%2B+Cos%28x%29+%2B+2+=+0

6Cos%5E2x-3+%2B+Cos%28x%29+%2B+2+=+0

6Cos%5E2x+%2B+Cos%28x%29+-+1+=+0

Factor the left side:

%283Cos%28x%29-1%29%282Cos%28x%29%2B1%29=0

Use the zero-factor principle for the first factor:

3Cos%28x%29-1=0

3Cos%28x%29=1

Cos%28x%29+=+1%2F3

We find the inverse cosine of abs%281%2F3%29 as 70.52877937°, which is
the reference angle.

Rounded to the nearest degree is 71°.  Since 1%2F3 is a positive
number and the Cosine is positive in the 1st and 4th quadrants, the
solutions are 71° for the 1st quadrant solution (same as the reference
angle) and 360°-70.52877937° = 289.4712206 or 289° for the 4th quadrant 
solution.


%283Cos%28x%29-1%29%282Cos%28x%29%2B1%29=0

Use the zero-factor principle for the second factor:

2Cos%28x%29%2B1=0

2Cos%28x%29=-1

Cos%28x%29+=+-1%2F2

We find the inverse cosine of abs%28-1%2F2%29 as 60°, the reference
angle.

Since -1%2F2 is a negative number and the Cosine is negative
in the 2nd and 3rd quadrants, the solutions are 180°-60° = 120° 
for the 2nd quadrant solution and 180°+60° = 240° for the 3rd quadrant
solution.

So the four correct answers to the nearest degree are:

71°, 289°, 120°, 240°

Edwin