Solve gamma(t)=2sin(t)i+4cos(t)j+tkt=0

$$\gamma ( t ) = 2 \sin ( t ) i + 4 \cos ( t ) j + t k t = 0$$

Solve for j

$$\left\{\begin{matrix}j=-\frac{2i\sin(t)+kt^{2}}{4\cos(t)}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }t=\pi n_{1}+\frac{\pi }{2}\\j\in \mathrm{C}\text{, }&k=-\frac{2i\sin(t)}{t^{2}}\text{ and }\exists n_{1}\in \mathrm{Z}\text{ : }t=\frac{\pi \left(2n_{1}+1\right)}{2}\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply t and t to get t^{2}.

$$2i\sin(t)+4\cos(t)j+t^{2}k=0$$

Subtract 2i\sin(t) from both sides. Anything subtracted from zero gives its negation.

$$4\cos(t)j+t^{2}k=-2i\sin(t)$$

Subtract t^{2}k from both sides.

$$4\cos(t)j=-2i\sin(t)-t^{2}k$$

The equation is in standard form.

$$4\cos(t)j=-2i\sin(t)-kt^{2}$$

Divide both sides by 4\cos(t).

$$\frac{4\cos(t)j}{4\cos(t)}=\frac{-2i\sin(t)-kt^{2}}{4\cos(t)}$$

Dividing by 4\cos(t) undoes the multiplication by 4\cos(t).

$$j=\frac{-2i\sin(t)-kt^{2}}{4\cos(t)}$$

Divide -2i\sin(t)-t^{2}k by 4\cos(t).

$$j=-\frac{2i\sin(t)+kt^{2}}{4\cos(t)}$$

Solve for k

$$\left\{\begin{matrix}k=-\frac{2\left(2j\cos(t)+i\sin(t)\right)}{t^{2}}\text{, }&t\neq 0\\k\in \mathrm{C}\text{, }&t=0\text{ and }j=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply t and t to get t^{2}.

$$2i\sin(t)+4\cos(t)j+t^{2}k=0$$

Subtract 2i\sin(t) from both sides. Anything subtracted from zero gives its negation.

$$4\cos(t)j+t^{2}k=-2i\sin(t)$$

Subtract 4\cos(t)j from both sides.

$$t^{2}k=-2i\sin(t)-4\cos(t)j$$

The equation is in standard form.

$$t^{2}k=-4j\cos(t)-2i\sin(t)$$

Divide both sides by t^{2}.

$$\frac{t^{2}k}{t^{2}}=\frac{2\left(-2j\cos(t)-i\sin(t)\right)}{t^{2}}$$

Dividing by t^{2} undoes the multiplication by t^{2}.

$$k=\frac{2\left(-2j\cos(t)-i\sin(t)\right)}{t^{2}}$$

Divide 2\left(-i\sin(t)-2\cos(t)j\right) by t^{2}.

$$k=-\frac{2\left(2j\cos(t)+i\sin(t)\right)}{t^{2}}$$