I think the correct answer is c
x² + y² = 9 is the equation of a circle with center (0, 0) and radius 3.
This puts the vertices at (-3, 0), (3, 0), (0, -3), and (0, 3).
When is the tangent line vertical? <em>when it passes through the x-axis.</em>
Answer: (-3, 0), (3, 0)
For this case, we must indicate which of the given functions is not defined for
By definition, we know that:
has a domain from 0 to infinity.
Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. For it to be defined, the term within the root must be positive.
Thus, we observe that:
is not defined, the term inside the root is negative when
.
While
if it is defined for 
, your domain is given by all real numbers.
Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. In the same way, its domain will be given by the real numbers, independently of the sign of the term inside the root.
So, we have:
with x = 0:
is defined.
with x = 0:
in the same way is defined.
Answer:

Option b
![\bf sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta) \\\\\\ tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}\\\\ -----------------------------\\\\ 2cos(A)=3tan(A)\implies 2cos(A)=3\cfrac{sin(A)}{cos(A)} \\\\\\ 2cos^2(A)=3sin(A)\implies 2[1-sin^2(A)]=3sin(A) \\\\\\ 2-2sin^2(A)=3sin(A)\implies 2sin^2(A)+3sin(A)-2](https://tex.z-dn.net/?f=%5Cbf%20sin%5E2%28%5Ctheta%29%2Bcos%5E2%28%5Ctheta%29%3D1%5Cimplies%20cos%5E2%28%5Ctheta%29%3D1-sin%5E2%28%5Ctheta%29%0A%5C%5C%5C%5C%5C%5C%0Atan%28%5Ctheta%29%3D%5Ccfrac%7Bsin%28%5Ctheta%29%7D%7Bcos%28%5Ctheta%29%7D%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0A%0A2cos%28A%29%3D3tan%28A%29%5Cimplies%202cos%28A%29%3D3%5Ccfrac%7Bsin%28A%29%7D%7Bcos%28A%29%7D%0A%5C%5C%5C%5C%5C%5C%0A2cos%5E2%28A%29%3D3sin%28A%29%5Cimplies%202%5B1-sin%5E2%28A%29%5D%3D3sin%28A%29%0A%5C%5C%5C%5C%5C%5C%0A2-2sin%5E2%28A%29%3D3sin%28A%29%5Cimplies%202sin%5E2%28A%29%2B3sin%28A%29-2)
![\bf \\\\\\ 0=[2sin(A)-1][sin(A)+2]\implies \begin{cases} 0=2sin(A)-1\\ 1=2sin(A)\\ \frac{1}{2}=sin(A)\\\\ sin^{-1}\left( \frac{1}{2} \right)=\measuredangle A\\\\ \frac{\pi }{6},\frac{5\pi }{6}\\ ----------\\ 0=sin(A)+2\\ -2=sin(A) \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%5C%5C%5C%5C%5C%5C%0A0%3D%5B2sin%28A%29-1%5D%5Bsin%28A%29%2B2%5D%5Cimplies%20%0A%5Cbegin%7Bcases%7D%0A0%3D2sin%28A%29-1%5C%5C%0A1%3D2sin%28A%29%5C%5C%0A%5Cfrac%7B1%7D%7B2%7D%3Dsin%28A%29%5C%5C%5C%5C%0Asin%5E%7B-1%7D%5Cleft%28%20%5Cfrac%7B1%7D%7B2%7D%20%5Cright%29%3D%5Cmeasuredangle%20A%5C%5C%5C%5C%0A%5Cfrac%7B%5Cpi%20%7D%7B6%7D%2C%5Cfrac%7B5%5Cpi%20%7D%7B6%7D%5C%5C%0A----------%5C%5C%0A0%3Dsin%28A%29%2B2%5C%5C%0A-2%3Dsin%28A%29%0A%5Cend%7Bcases%7D)
now, as far as the second case....well, sine of anything is within the range of -1 or 1, so -1 < sin(A) < 1
now, we have -2 = sin(A), which simply is out of range for a valid sine, so there's no angle with such sine
so, only the first case are the valid angles for A