Question

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

Answer 1

`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`.