How do I solve #sinx-sqrt3sin3x+sin5x<0#?

1 Answer
Nghi N.
Jun 13, 2018

#sin x + sin (5x) - sqrt3sin(3x)< 0# (1)
Reminder.
#sin x + sin 5x = 2sin (3x).cos (2x)#
Equation (1) becomes:
#f(x) = 2sin (3x)cos (2x) - sqrt3sin (3x) = #
#f(x) = sin (3x)(2cos 2x - sqrt3)< 0# (2)
First, solve the 2 basic trig equations to find all the end-points (critical points).
#g(x) = sin 3x = 0# --> x = 0, #x = pi#, and #x = 2pi#
#h(x) = 2cos 2x - sqrt3 = 0# --> #cos 2x = sqrt3/2#
Trig table and unit circle -->
#2x = +- pi/6# --> #x = +- pi/12#
#x = pi/12#, and #x = 2pi - pi/12 = (23pi)/12# (co-terminal).
To algebraically solve the trig inequality (1) we have to create a sign chart (or sign table). This page format doesn't allow me to create a sign chart. So, you have to find the method shown in trig books. You also can solve the inequality (2) by graphing calculator. The parts of the graph, that are below the x-axis represent the answers by intervals. See book titled:"Solving trig equations and inequalities" (Amazon Kindle).

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