Solve for Sin teta=0 | Math Homework Help

[lioric]

Homework Statement

Homework Equations

The Attempt at a Solution

The first picture is the question
The second picture is the marking scheme
I have circled in yellow the problem
I would like to know how sin teta = 0
Thank you

 

[PeroK]

I would like to know how sin teta = 0
Thank you

What don't you understand about it?

 

[lioric]

What don't you understand about it?

I do know how cos teta = 1/3
But the sin teta part I don't get I couldn't transpose the formula 3sin teta = sin teta / cos teta to sin teta = 0
So I know I m missing something fundamental

 

[PeroK]

I do know how cos teta = 1/3
But the sin teta part I don't get I couldn't transpose the formula 3sin teta = sin teta / cos teta to sin teta = 0
So I know I m missing something fundamental

What is $\tan(0)$?

What is $3\sin(0)$?

 

[lioric]

What is $\tan(0)$?

What is $3\sin(0)$?

Both is 0

 

[PeroK]

Both is 0

So, $\theta = 0$ is a solution to your equation $\tan \theta = 3\sin \theta$.

What about other values of $\theta$ where $\sin \theta = 0$?

 

[lioric]

So, $\theta = 0$ is a solution to your equation $\tan \theta = 3\sin \theta$.

What about other values of $\theta$ where $\sin \theta = 0$?

Hmmm I guess I know what your saying
Teta= 180
That part I understand
Thank you very much

 

[Ray Vickson]

Homework Statement

View attachment 109209
View attachment 109210

Homework Equations

The Attempt at a Solution

The first picture is the question
The second picture is the marking scheme
I have circled in yellow the problem
I would like to know how sin teta = 0
Thank you

The book's explanation is poorly worded; it should have said $\cos(\theta) = 1/3$ OR $\sin(\theta) = 0$. That means that one or the other of those two possibilities can occur, and that corresponds to the existence of more than one solution to the problem in the interval $0^o < \theta < 360^o$.

If you plot the graphs of $y = 3\sin(\theta)$ and $y = \tan(\theta)$ over the interval $0^o < \theta < 360^o$, you will see that there are three points where the two graphs cross (five points of crossing if you include $\theta = 0^o$ and $\theta =360^o$). The equation $\sin(\theta) = 0$ has one solution in $(0^0,360^o)$, while the equation $\cos(\theta) = 1/3$ has two solutions in $(0^o,360^o)$.

 

[lioric]

The book's explanation is poorly worded; it should have said $\cos(\theta) = 1/3$ OR $\sin(\theta) = 0$. That means that one or the other of those two possibilities can occur, and that corresponds to the existence of more than one solution to the problem in the interval $0^o < \theta < 360^o$.

If you plot the graphs of $y = 3\sin(\theta)$ and $y = \tan(\theta)$ over the interval $0^o < \theta < 360^o$, you will see that there are three points where the two graphs cross (five points of crossing if you include $\theta = 0^o$ and $\theta =360^o$). The equation $\sin(\theta) = 0$ has one solution in $(0^0,360^o)$, while the equation $\cos(\theta) = 1/3$ has two solutions in $(0^o,360^o)$.

This is from an alevel past paper
But this explanation is very nice
Thank you very much