pedzenekO

2021-08-20

Find all solution of the equation in the interval $\left[0,2\pi \right]$.
$\left(\mathrm{cos}x-1\right)\left(\mathrm{sin}x+1\right)=0$

Layton

$\left(\mathrm{cos}x-1\right)\left(\mathrm{sin}x+1\right)=0$
The given equation's solution, i.e., must now be found.
The entire expression will be equal to 0 if any one factor on the left side of the equation is equal to 0.
$\mathrm{cos}x-1=0$
$\mathrm{sin}x-1=0$
Set the first factor equal to 0
$\mathrm{cos}x-1=0$
$\mathrm{cos}x=1$
Take the inverse cosine of both sides of the equation to extract x from inside the cosine
$x=\mathrm{arccos}\left(1\right)$
The exact value of $\mathrm{arccos}\left(1\right)$ is 0
$x=2\pi -0$
Similarly $\mathrm{sin}\left(x\right)=-1$
Taking the inverse sine of both sides of the equation to exact x from inside the sine
$x=\mathrm{arcsin}\left(-1\right)$
The exact value of $\mathrm{arcsin}\left(-1\right)$ is $\frac{-\pi }{2}$
$x=\frac{-\pi }{2}$ but it should be between $\left[0,-2\pi \right]$
$x=2\pi +\left(-\frac{\pi }{2}\right)=\frac{3\pi }{2}$
$x=\frac{3\pi }{2}$

Jeffrey Jordon