How do you solve #4cosx+3=2/cosx# in the interval #0<=x<=2pi#?

1 Answer
Nghi N
Feb 23, 2017

#63^@26; 296^@74#

Explanation:

Cross multiply, then bring equation to standard form:
#4cos^2 x + 3cosx - 2 = 0#
Solve this quadratic equation by the improved quadratic formula (Socratic Search)
#D = d^2 = b^2 - 4ac = 9 + 32 = 41# --> #d = +- sqrt41#
There are 2 real toots:
#cos x = - 3/8 +- sqrt41/8 = (-3 +- sqrt41)/8#
#cos x1 = (-3 + sqrt41)/8 = 0.45#
#cos x2 = (-3 - sqrt41)/8 = - 1.18# (Rejected as < -1)
Calculator and unit circle -->
cos x = 0.45 --> #x = +- 63^@26#
arc (- 63.26) is co-terminal to (360 - 63.26 = 296^@74)
Answers for (0, 360):
#63^@26; 296^@74#

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