3𝑠𝑖𝑛2𝑥 = 2𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥 is this a provable? if not, solve the equation for -180 < 𝑥 < 180?

First, we may notice that #color(red)(2sin(x)cos(x))# is equal to #color(red)sin(2x)# by the double angle formula for the sine function.

#3sin(2x) = color(red)(2sin(x)cos(x))#

#3sin(2x) = sin(2x)#

Obviously, these two are not equivalent. We must then solve for #-180 < x < 180#.

To make it a little bit easier to understand what's going on, we may want to make a substitution for the trigonometric function part of our equation.

Let #color(blue)u = color(red)(sin(2x))#

#3color(blue)u = color(blue)u#

#2color(blue)u = 0#

#color(blue)u = 0#

#color(red)sin(2x) = 0#

Instinctually, you may want to take the arc sine of both sides. Just remember that because sine is a periodic function, we will definitely have more than one possible #2x# for which the equation holds true.

#2x = 0#
#color(green)(x = 0)#

But we can add #color(purple)(180n)#, where #n# is some integer, to get more solutions.

#2x = 180#
#color(green)(x = 90)#

#2x = -180#
#color(green)(x = -90)#

Note that other values for #x#, such as #x = 180#, do work in the equation, but do not follow the inequalities #-180 < x < 180#.

So, the possible values for #x# in the equation #3sin(2x) = 2sin(x)cos(x)# are #x = 0#, #x = 90#, and #x = -90#.