iarc6io

2022-07-25

Solve the equation on the interval $\left[0,2\pi \right]$
$\mathrm{sin}2x-\mathrm{sin}x=0$

Hassan Watkins

Expert

$\mathrm{sin}2x-\mathrm{sin}x=0$
The first thing to do is to factor out sinx. We take out sinxand divide sin2x and sinx by what has been taken out or sinx.
Doing the algebra, we get this trig equation:
$\mathrm{sin}x\left(\mathrm{sin}x-1\right)=0$
We have two factors:
sinx and sinx - 1
Set each factor to 0 and solve for sinx just like you did backin algebra 1.
sinx = 0
When does sinx = 0 between 0 degrees and $2\pi$?
sinx is zero at 0 degrees, 180 degrees and 360 degrees. By the way, to say that sinx is zero at 0 degrees and 360degrees means the samething.
We now set the other factor to zero.
sinx - 1 = 0
Add one to both sides just like you did in algebra 1.
sinx = 1
When does sinx = 1?
Sinx = 1 at 90 degrees.
All degrees are: 0 degrees (the same as 360 degrees), 180degrees and 90 degrees.
All radian measures are: $\pi ,2\pi$ and $\frac{\pi }{2}$