SOLUTION: {{{ g(x) = log_3((x+6)/2) }}} i. find the domain of g in interval form ii. find the x-intercept(s) of the graph of g (iii) In what quadrant does the terminal side of an ang

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Question 969216:
i. find the domain of g in interval form
ii. find the x-intercept(s) of the graph of g

(iii) In what quadrant does the terminal side of an angle of 5 radians lie? justify your answer
(iv) Find all solutions for the equation:
Thank you
Found 3 solutions by solver91311, lwsshak3, ikleyn:
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!

The beginning of Quadrant IV is which is a little over 4 and a half. The end of Quadrant IV is which is a little over 6 and a quarter.

John

My calculator said it, I believe it, that settles it

Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website!

i. find the domain of g in interval form
x+6≥0
x≥-6
(-6, ∞)
..
ii. find the x-intercept(s) of the graph of g
set g=0
log_3((x+6)/2)=0
exponential form: base(3) raised to log of number(0)=number((x+6)/2)
3^0=(x+6)/2
1=(x+6)/2
x+6=2
x=-4
..
In what quadrant does the terminal side of an angle of 5 radians lie? justify your answer
use degrees:
5 radians/π*180≈286˚
5 radians lie in quadrant IV
..
Find all solutions for the equation:

sec(4t)=-2/√3
cos(4t)=-√3/2
4t=5π/6+2πk, 7π/6+2πk, k=any integer
t=5π/24+2πk, 7π/24+2πk, k=any integer

Answer by ikleyn(53875)   (Show Source):
You can put this solution on YOUR website!
.
iv. Find all solutions for the equation:
~~~~~~~~~~~~~~~~~~~~~~~~~~~

        The solution to this problem (iv) in the post by @lwsshak3 is fundamentally (conceptually) wrong.

        I will copy-paste his solution here, will show his error and will write a correct/(the corrected) version.

sec(4t) = cos(4t) = 4t = , , k = any integer t = , , k = any integer <<<---=== this is C O R R E C T I O N t = , , k = any integer <<<---=== this is CORRECT

Solved correctly.