How do you solve #sin2x=-sinx#?

1 Answer
Nghi N.
Apr 30, 2015

Solve #f(x) = sin 2x + sin x = 0 #.
Use the trig identity:# sin 2x = 2.sinx.cos x#

f(x) = sin x(2.cos x - 1) = 0.

Next solve the 2 basic trig equation: #sin x = 0 and 2.cos x = 1.#

#sin x = 0 --> x = 0; and x = 2Pi#

#cos x = 1/2 --> trig conversion table gives -> cos x = cos (Pi/3) --> x = Pi/3#.
Trig unit circle gives another arc that has the same cos value (1/2):
#x = (5Pi)/3.#
#Within interval (0, 2Pi): there are 4 answers: 0; Pi/3; 5Pi/3; and 2Pi#.

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