question 4 solve for t, 0 ≤ t < 2π. 27 sin(t) cos(t) = -9 sin(t) give your answers separated by commas…

Explanation:

Step1: Rearrange the equation

Move all terms to one - side: $27\sin(t)\cos(t)+9\sin(t) = 0$. Then factor out $\sin(t)$: $\sin(t)(27\cos(t)+9)=0$.

Step2: Solve $\sin(t)=0$

We know that $\sin(t)=0$ when $t = k\pi$, $k\in\mathbb{Z}$. Since $0\leq t<2\pi$, then $t = 0,\pi$.

Step3: Solve $27\cos(t)+9 = 0$

First, isolate $\cos(t)$: $27\cos(t)=-9$, so $\cos(t)=-\frac{9}{27}=-\frac{1}{3}$. Then $t=\arccos(-\frac{1}{3})$. Using a calculator, $t\approx1.91$ (in the second - quadrant) and $t = 2\pi-\arccos(-\frac{1}{3})\approx4.37$ (in the third - quadrant).

Answer:

$0,1.91,\pi,4.37$