Question

How do you solve `4tanx + sin2x = 0`?

Answer 1

Any integer multiple of `\pi`, [ ex) `x=0,\pi,-pi,2pi,-2pi, ... ]` will solve the equation.

Use that `tanx=sinx/cosx` and that `(sin2x=2sinx cosx)`

Making these substitutions,

`4tanx+sin2x=4sinx/cosx+2sinxcosx=`

`=sinx(4/cosx+2cosx)`

One way this expression can equal zero is if `sin x=0`. `sin x` will be zero whenever `x` is an integer multiple of `pi`.

Are there any other solutions?

The expression will also be zero when `4/cosx+2cosx=0`.

The preceding equation can be rearranged into the form,

`(cosx)^2=-2`

Because the square of a real number cannot be negative there are no solutions to this equation.

Therefore, integer multiples of `pi` are the only solutions.

Answer 2

There is another way that gives same answer.
Call tan x = t . Use trig identity: `sin 2x = (2t)/(1 + t^2)`

`f(x) = 4t + (2t)/(1 + t^2) = [(2t).(3 + 2t^2)]/(1 + t^2)` = 0
The 2 quantities with t^2 are always positive.
f(x) = 0 when t = tan x = 0 -> x = 0 or `x = K.Pi`