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

1 Answer
Feb 25, 2018

3sin(2x) = 2sin(x)cos(x) is not provable. The possible values for x in the equation 3sin(2x) = 2sin(x)cos(x), such that -180 < x < 180, are x = 0, x = 90, and x = -90.

Explanation:

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.