Solve operatorname{asin}xcosx+sinx=0

$$\operatorname { asin } x \cos x + \sin x = 0$$

Solve for a

$$\left\{\begin{matrix}a=-\frac{1}{\cos(x)}\text{, }&\nexists n_{2}\in \mathrm{Z}\text{ : }x=\pi n_{2}+\frac{\pi }{2}\\a\in \mathrm{R}\text{, }&\exists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}\end{matrix}\right.$$

Steps for Solving Linear Equation

Subtract \sin(x) from both sides. Anything subtracted from zero gives its negation.

$$a\sin(x)\cos(x)=-\sin(x)$$

The equation is in standard form.

$$\frac{1}{2}\sin(2x)a=-\sin(x)$$

Divide both sides by \sin(x)\cos(x).

$$\frac{\frac{1}{2}\sin(2x)a}{\frac{1}{2}\sin(2x)}=-\frac{\sin(x)}{\frac{1}{2}\sin(2x)}$$

Dividing by \sin(x)\cos(x) undoes the multiplication by \sin(x)\cos(x).

$$a=-\frac{\sin(x)}{\frac{1}{2}\sin(2x)}$$

Divide -\sin(x) by \sin(x)\cos(x).

$$a=-\frac{1}{\cos(x)}$$

Solve for x

$$\left\{\begin{matrix}\\x=\pi n_{1}\text{, }n_{1}\in \mathrm{Z}\text{, }&\text{unconditionally}\\x=\arccos(\frac{1}{a})+2\pi n_{2}-\pi \text{, }n_{2}\in \mathrm{Z}\text{; }x=-\arccos(\frac{1}{a})+2\pi n_{3}+\pi \text{, }n_{3}\in \mathrm{Z}\text{, }&|a|\geq 1\end{matrix}\right.$$