Solve r(t)=cos(t)i+5tj+4sin(3t)kquadt=0

$$r ( t ) = \cos ( t ) i + 5 t j + 4 \sin ( 3 t ) k \quad t = 0$$

Solve for j

$$j=\frac{-4k\sin(3t)-\frac{i\cos(t)}{t}}{5}$$

$$t\neq 0$$

Steps for Solving Linear Equation

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

$$5tj+4\sin(3t)kt=-i\cos(t)$$

Subtract 4\sin(3t)kt from both sides.

$$5tj=-i\cos(t)-4\sin(3t)kt$$

The equation is in standard form.

$$5tj=-4kt\sin(3t)-i\cos(t)$$

Divide both sides by 5t.

$$\frac{5tj}{5t}=\frac{-4kt\sin(3t)-i\cos(t)}{5t}$$

Dividing by 5t undoes the multiplication by 5t.

$$j=\frac{-4kt\sin(3t)-i\cos(t)}{5t}$$

Divide -i\cos(t)-4kt\sin(3t) by 5t.

$$j=\frac{-4k\sin(3t)-\frac{i\cos(t)}{t}}{5}$$

Solve for k

$$\left\{\begin{matrix}k=-\frac{i\cos(t)+5jt}{4t\sin(3t)}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }t=\frac{\pi n_{1}}{3}\\k\in \mathrm{C}\text{, }&j=-\frac{i\cos(t)}{5t}\text{ and }t\neq 0\text{ and }\exists n_{1}\in \mathrm{Z}\text{ : }t=\frac{\pi n_{1}}{3}\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$5tj+4\sin(3t)kt=-i\cos(t)$$

Subtract 5tj from both sides.

$$4\sin(3t)kt=-i\cos(t)-5tj$$

The equation is in standard form.

$$4t\sin(3t)k=-i\cos(t)-5jt$$

Divide both sides by 4\sin(3t)t.

$$\frac{4t\sin(3t)k}{4t\sin(3t)}=\frac{-i\cos(t)-5jt}{4t\sin(3t)}$$

Dividing by 4\sin(3t)t undoes the multiplication by 4\sin(3t)t.

$$k=\frac{-i\cos(t)-5jt}{4t\sin(3t)}$$

Divide -i\cos(t)-5tj by 4\sin(3t)t.

$$k=-\frac{\frac{i\cos(t)}{t}+5j}{4\sin(t)\left(4\left(\cos(t)\right)^{2}-1\right)}$$