How do you find all solutions of the following equation in the interval [0,2pi] of 3cos2x+cosx+2=0?

1 Answer
Nghi N.
Apr 8, 2016

70^@53; 289^@47, 60^@; 300^@

Explanation:

3cos 2x + cos x + 2 = 0
Replace cos 2x by (2cos^2 x - 1) -->
3(2cos^2 x - 1) + cos x + 2 = 0
6cos^2 x - 3 + cos x + 2 = 0
Solve this quadratic equation for cos x
6cos^2 x + cos x - 1 = 0
D = d^2 = b^2 - 4ac = 1 + 24 = 25 --> d = +- 5
There are 2 real roots:
cos x = - b/(2a) +- d/(2a) = -1/12 +- 5/12
cos x = - 1/12 + 5/12 = 4/12 = 1/3
cos x = - 1/12 - 5/12 = - 6/12 = -1/2
a. cos x = 1/3 --> x = +- 70^@53
Co-terminal arc of x = - 70.53 --> x = 289^@47
b. cos x = -1/2 ---> x = +- 60^@
Co-terminal of x = - 60 ---> x = 300^@
Answers for (0, 2pi):
70^@53; 289^@47; 60^@; 300^@

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