How do you solve $sin x cos x + cos x = 0$ $sinxcosx+cosx=0$ ?

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Answer 1 · 2016-09-25T10:02:27

$x = n π \pm \frac{π}{2}$ $x=npi+-pi/2$ , where $n$ $n$ is an integer

$sin x cos x + cos x = 0$ $sinxcosx+cosx=0$

$\Leftrightarrow cos x (sin x + 1) = 0$ $hArrcosx(sinx+1)=0$

i.e. either $cos x = 0$ $cosx=0$ and $x = (n π \pm \frac{π}{2})$ $x=(npi+-pi/2)$

or $sin x = - 1$ $sinx=-1$ and $x = 2 n π - \frac{π}{2}$ $x=2npi-pi/2$

where $n$ $n$ is an integer

As the two are contained in $x = (n π \pm \frac{π}{2})$ $x=(npi+-pi/2)$ , where $n$ $n$ is an integer, this is the solution.

How do you solve sinxcosx+cosx=0sinxcosx+cosx=0sinxcosx+cosx=0?