Question

Math: https://snipboard.io/4jncTZ.jpg

Answer 1

@oliver69

Answer 2

a.
\[y=2\cos \theta+\sqrt{3}\]
when it is in original position y=0
\[0=2\cos \theta+\sqrt{3}\]
\[\cos \theta=-\frac{ \sqrt{3} }{ 2 }=-\cos \frac{ \pi }{ 6 }=\cos (\pi-\frac{ \pi }{ 6 }),\cos (\pi+\frac{ \pi }{ 6})\]
\[\theta=\pi-\frac{ \pi }{ 6 },\pi+\frac{ \pi }{ 6 }=\frac{ 5\pi }{ 6 },\frac{ 7\pi }{ 6}\]
\[\theta=2k \pi+\frac{ 5\pi }{ 6 },2k \pi+\frac{ 7\pi }{ 6 }\]
where k is a whole number.
b.
\[y=2\cos 2\theta+\sqrt{3}\]
when y=0
\[0=2\cos 2\theta+\sqrt{3}\]
\[\cos 2\theta=-\frac{ \sqrt{3} }{ 2 }=-\cos \frac{ \pi }{ 6 }=\cos (\pi-\frac{ \pi }{ 6 }),\cos (\pi+\frac{ \pi }{ 6 })\]
\[=\cos (2k \pi+\frac{ 5\pi }{ 6}),\cos (2k \pi+\frac{ 7\pi }{ 6 }),k=0,1\]
\[2\theta =2k \pi+\frac{ 5\pi }{ 6 },2k \pi+\frac{ 7\pi }{ 6 },k=0,1\]
\[2\theta=\frac{ 5\pi }{ 6 },\frac{ 7\pi }{ 6 },\frac{ 17\pi }{ 6},\frac{ 19\pi }{ 6}\]
\[\theta=\frac{ 5\pi }{ 12 },\frac{ 7\pi }{ 12 },\frac{ 17\pi }{ 12 },\frac{ 19\pi }{ 12}\]

Answer 3

Wow this is just great thank you

Answer 4

And this is part C.) https://snipboard.io/zL6EpX.jpg

Answer 5

\[y=2\cos \theta+1~and ~y=\sin^2\theta-1\]
lengths are equal when
\[2\cos \theta+1=\sin^2\theta-1=(1-\cos ^2\theta)-1=-\cos ^2\theta \]
or
\[\cos ^2\theta+2\cos \theta+1=0,(\cos \theta+1)^2=0\]
\[\cos \theta=-1=cos \pi=\cos(2k\pi+\pi)\]
\[\theta=\pi+2k \pi \]
where k is a whole number.

Answer 6

Omg thank god you are online i need help with like 5 other questions please

Answer 7

sorry i am going to bed,tomorrow

Answer 8

Awwww that suck okay thanks