How do you find the exact value of #3cos2theta+costheta+2=0# in the interval #0<=theta<2pi#?

1 Answer
Jul 13, 2017

#x = +- 120^@#
#x = +- 70^@53#

Explanation:

Replace cos 2t by #(2cos^2 t - 1)# -->
#3(2cos^2 t - 1) + cos t + 2 = 0#
Solve this quadratic equation for cos t
#6cos^2 t + cos t - 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 = (-1 +- 5)/12#
#cos x = - 6/12 = -1/2#
#cos x = 4/12 = 1/3#
Use calculator and unit circle:
a. #cos x = - 1/2# --> #x = +- 120^@#
b. #cos x = 1/3# --> #x = +- 70^@53#