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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