Question 1195187
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Part 1: Finding the domain.


Recall that the smallest cosine can get is -1, and the largest is 1
I.e. the outputs are between -1 and 1, inclusive of both endpoints


This tells us the range is
{{{-1 <= cos(x) <= 1}}}


This then leads directly to the fact that the domain of *[tex \Large \cos^{-1}(x)] is {{{-1 <= x <= 1}}}. 
The domain and range swap places when going from the original to the inverse function.
This swap is because x and y themselves swap roles. 


Therefore, the input expression 3x+2 must be between -1 and 1
{{{-1 <= 3x+2 <= 1}}}


{{{-1-2 <= 3x+2-2 <= 1-2}}} Subtracting 2 from all sides


{{{-3 <= 3x <= -1}}}


{{{-3/3 <= 3x/3 <= -1/3}}} Dividing all sides by 3


{{{-1 <= x <= -1/3}}}
The inequality signs stay the same the entire time. They only would flip if we divided all sides by a negative number.


Answer: The domain as a compound inequality is  {{{-1 <= x <= -1/3}}}
Verbally we can say x must be between -1 and -1/3, inclusive of both endpoints.
In interval notation, the domain is [-1, -1/3]. Don't forget the square brackets.


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Part 2: Solve the equation for x when y = pi/4


*[tex \Large y = \cos^{-1}(3\text{x}+2)]


*[tex \Large \frac{\pi}{4} = \cos^{-1}(3\text{x}+2)]


*[tex \Large \cos\left(\frac{\pi}{4}\right) = 3\text{x}+2]


*[tex \Large \frac{\sqrt{2}}{2} = 3\text{x}+2]


*[tex \Large \frac{\sqrt{2}}{2}-2 = 3\text{x}]


*[tex \Large 3\text{x} = \frac{\sqrt{2}}{2}-2]


*[tex \Large 3\text{x} = \frac{\sqrt{2}}{2}-\frac{4}{2}]


*[tex \Large 3\text{x} = \frac{\sqrt{2}-4}{2}]


*[tex \Large \text{x} = \frac{\sqrt{2}-4}{2}*\frac{1}{3}]


*[tex \Large \text{x} = \frac{\sqrt{2}-4}{6}]
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