Question

What are the values that satisfy the trigonometric equation for: 0 < q < 2?

Sinθ + tan(-θ) = 0
A.
B.
C.
D.

Answer 1

Okay. So lets do this by hand this time :)

Answer 2

So, this is what you have: \(\sin (\emptyset) + \tan (-\emptyset) = 0 \)

Answer 3

ok how would we do that?

Answer 4

Okay. Now we are going to replace tan just to give you the idea:

\(\large \sin (\emptyset) + \frac{ \sin (-\emptyset) }{ \cos (-\emptyset) } = 0 \)

Answer 5

Now make sure that you realize: \(\color{blue} \sin\color{blue}(\color{blue}-\color{blue}x\color{blue}) \color{blue}=\color{blue}s\color{blue}i\color{blue}n\color{blue} (\color{blue}x) \) and \(\color{blue}c\color{blue}o\color{blue}s\color{blue}(\color{blue}-\color{blue}x\color{blue})\color{blue} = \color{blue}c\color{blue}o\color{blue}s\color{blue}(\color{blue}x\color{blue})\)

Answer 6

ok so the negatives = positives

Answer 7

Yes! :) So, cosine is 1, and sine is 0

These are the values of the functions. All the angles are AT, where sine is 0, and also when cosine is 1.

Now, this is where the unit circle comes in. Check your unit circle to find out where, sine = 0 and cosine = 1 is.

UNIT CIRCLE: http://i1.cpcache.com/product_zoom/627420071/large_unit_circle_clock.jpg

Answer 8

Well, can you give me the answer choices? :)

Answer 9

\[A. \frac{ \pi }{ 4 }, \frac{ 5\pi }{ 4 }\]

\[B. 0, \frac{ \pi }{ 2 }, \pi, \frac{ 3\pi }{ 2 }\]

\[C. 0, \pi\]

\[D. \frac{ \pi }{ 2 },\frac{ 3\pi }{ 2 }\]

Answer 10

So, for the rest of the work... I am going to use \(x\) in place of the theta.

So the equation that was given to use will look like: \(sin(x)+tan(-x)=0 \)

Now, it will become \(sin(x)-tan(x)=0\). This is because \(tan(x)\) is odd.

So, we are going to replace tan just like we did before: \(sin(x)-sin(x)/cos(x)=0 \)

Now it will turn into: \(sin(x)*cos(x)/cos(x)-sin(x)/cos(x)=0 \)

So just continuing to solve: \(( sin(x)*cos(x)-sin(x) )/cos(x)=0 \)

\(sin(x)*cos(x)-sin(x) = 0*cos(x) \)

\(sin(x)*cos(x)-sin(x) = 0 \)

\(sin(x)( cos(x)-1 ) = 0 \)

No it can either be \(sin(x) = 0\) or it can be \(cos(x)-1 = 0 \)... either thing.

So it is also: \(sin(x) = 0\) or \(cos(x) = 1 \)

Now we just solved it a bit... \(x = arcsin(0)\) or \(x = arccos(1) \)

So this means that \(x = 0\), \(pi\) or \(x = 0 \)


Now. After all that work... if \(x\) is in the interval, which is \([0, 2pi]\), that means that the solutions are: \(x = 0\) and \(x = pi\).

Answer 11

So the final answer is \(x = 0\) and \(x = pi\)... C! :)

Answer 12

Hope this helped! Have a great day! :)

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Answer 13

THANK YOU!!!!!!!!!!!!!!!!!