How do you solve #2cos^2(x/3) + 3sin(x/3) - 3 =0#?

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Answer 1 · 2015-05-06T16:09:12

Cal #sin (x/3) = t.#
#y = 2(1 - t^2) + 3t - 3 = 0#
Quadratic equation.

# -2t^2 + 3t - 1 = 0#

Since a + b + c = 0, one real root is t= 1 and the other is (t = 1/2)

a. #t = sin (x/3) = 1 -> x/3 = pi/2 -> x = (3pi)/2#

b. #sin (x/3) = t = 1/2 -> x/3 = pi/6 -> x = (3pi)/6 = pi/2#

c. #sin (x/3) = 1/2 -> x/3 = (5pi)/6 -> x = (15pi)/6 = (pi/2 + 2pi).#

Within interval (0, 2pi), there are 2 answers:# pi/2 and (3pi)/2.#

Always check the answers.

#x = pi/2 -> x/3 = pi/6 -> y = 2.(3/4) + 3.(1/2) - 3 = 3/2 + 3/2 - 3 = 0#. Correct
#x = (3pi)/2 -> x/3 = pi/2 -> y = 0 + 3 - 3 = 0# Correct.