Question

How do you solve `2cos(x)sin(x)+sin(x)=0`?

Answer 1

`x= 0^o or 120^o or 180^o or 240^o`
or
`x= 0 or (2pi/3) or pi or (4pi)/3` radians
`color(white)("XXXX")`assuming we are restricted to `[0,360^o)`

Explanation:

If `2cos(x)sin(x)+ sin(x) = 0`
then
`color(white)("XXXX")` `(sin(x))*(2cos(x)+1) = 0`

`rArr sin(x) = 0`
`color(white)("XXXX")` `rarr x = 0^o or 180^o`
or
`rArr 2cos(x)+1 = 0`
`color(white)("XXXX")` `rarr cos(x) = -1/2`

`color(white)("XXXX")` `rarr x= 180^o +- 60^o`
`color(white)("XXXX")` `color(white)("XXXX")`(standard trigonometric triangle)
`color(white)("XXXX")` `rarr x = 120^o or 240^o`