How do you solve the equation #cos(x+pi/3)+cosx=0# on interval [0,2pi)?

1 Answer
Nghi N.
Mar 24, 2016

#pi/3 and (4pi)/3#

Explanation:

Apply the trig identity:
#cos a + cos b = 2cos ((a + b)/2).cos ((a - b)/2)#
#cos (x + pi/3) + cos x = 2cos (x + pi/6)cos (pi/6) = 0 #
Trig table --> cos pi/6 = sqrt3/2. The equation becomes:
#sqrt3(cos x + pi/6) = 0#
#cos (x + pi/6) = 0# --> 2 solutions:
a. #x + pi/6 = pi/2# --># x = pi/2 - pi/6 = (2pi)/6 = pi/3#
b. # x + pi/6 = (3pi)/2 # --># x = (3pi)/2 - pi/6 = (8pi)/6 = (4pi)/3#
Answers for #0, 2pi)#
#pi/3 and (4pi)/3#
Check.
#x = (4pi)/3 #--> #cos x = - cos pi/3 = -1/2.#
#cos (x + pi/3) = cos (5pi)/3 = cos pi/3 = 1/2#,
#cos x + cos (x + pi/3) = - 1/2 + 1/2 = 0.# OK

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