The cotangent graph only has a period ofintervals and is most similar to the tangent graph. The domain of cotangent exists everywhere except everyvalue since an asymptote exists at those values in the domain.

The y-intercept of 3 shifts the cotangent graph up by three units, so this does not affect the domain.

Therefore, the graph exists everywhere except, whereis an integer.

All x values make the function work. Thus, making the domain. They're parentheses instead of brackets because parentheses are used when you can't actually use the specific value next to it. It is impossible to use infinity which makes parentheses appropriate. Brackets are used when you CAN use the specific value next to it.

All x values work for the function. Thus, making the domain all real numbers. Parentheses are required because you can never actually use the number infinity.

If you look at a graph of the function, you can see that every curve has a vertical asymptote that repeats everyradians in the positive and negative x-direction, starting atradians. Also, the curve has a length that stretchesradians which makes the domain.

has restrictions on its domain such that

, where是任何整数。

To determine the domain for,

we equate the terms within the secant function and set them equal to the original domain restriction.

Solving for,

The new domain restriction is:

whereis an integer

Rewritein terms of sine and cosine functions.

Since these angles are special angles from the unit circle, the values of each term can be determined from the x and y coordinate points at the specified angle.

Solve each term and simplify the expression.

Using trigonometric relationships, one can set up the equation

.

Solving for,

Thus, the answer is found to be 29.

Using trigonometric relationships, one can set up the equation

.

Plugging in the values given in the picture we get the equation,

.

Solving for,

.

Thus, the answer is found to be 106.

The values ofthat fit this equation would be:

and

because these angles are in QI and QII where sin is positive and where

.

This is why the answer

is incorrect, because it includes inputs that provide negative values such as:

Thus the answer would be eachmultiple ofand, which would provide the following equations:

OR

To evaluate, break up each term into 3 parts and evaluate each term individually.

Simplify by combining the three terms.