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Find the domain: Y= log[sin(x-3)]+ √(16-x²)
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Find the domain: Y= log[sin(x-3)]+ √(16-x²)
Introduction
In order to find the domain of the given function, we need to consider the restrictions on the variables that could cause the function to be undefined or produce imaginary outputs.

Logarithmic Function
The function Y= log[sin(x-3)] involves a logarithmic function. The domain of a logarithmic function is restricted to positive real numbers. Therefore, sin(x-3) must be greater than zero.

Sine Function
The sine function sin(x) is defined for all real numbers. However, sin(x-3) must also be greater than zero. This restriction means that the value of x must lie within a certain range.

Radical Function
The function √(16-x²) involves a radical function. The domain of a radical function is restricted to non-negative real numbers. Therefore, 16-x² must be greater than or equal to zero.

Solving for x
To find the domain of the function, we need to solve for x. We begin by finding the values of x that make sin(x-3) equal to zero. This occurs when x-3 is equal to an odd multiple of π/2. Therefore, x=3+π/2+nπ, where n is an integer.

Next, we find the values of x that make 16-x² less than zero. This occurs when x is greater than 4 or less than -4. Therefore, the domain of the function is:

Domain: {x | -4 ≤ x ≤ 4, x ≠ 3+π/2+nπ, where n is an integer}
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Find the domain: Y= log[sin(x-3)]+ √(16-x²)
Given, 
     Y = log[sin(x-3)]+ √(16-x^2) = f(x)+ g(x);
     Let, f(x)= log[sin(x-3)] & g(x)= √(16-x^2); Domains of log() funcn & √() funcn are different.
     To find Y domain find f(x), g(x) domains it's intersection interval gives the answer i.e; Domain of Y.
     Finding domain of f(x) =log[sin(x-3)]  
          We know that domain of log x is (0,+∞). Implies sin(x-3) >0 & Domain of sin(x-3) is (-∞, +∞).But, now domain of sin(x-3) is (0, +∞). 0<(x-3)< ∞ . Now, add 3 to the inequality, it becomes  3<x<∞. Implies x>3 & x belongs to (3,∞).
          Domain of f(x)= log[sin(x-3)] is (3,∞).
    Finding domain of g(x)=√(16-x^2)
            We know that domain of a √(x) is [0, +∞). Implies x>= 0. Here, in √(16-x^2) function, (16-x^2)must be >=0. (4-x)(4+x)>=0. 
            If  (4-x)>=0 & (4+x)>=0  implies 4>=x(x<=4) &x>=-4 Now, x is [-4,4].
            Domain of g(x) is [-4,4].
     Now Resultant domain is common intervals of both f(x) &g(x). x belongs to (3,∞) & [-4,4]. It's common interval is (3,4].
           Therefore Domain of Y is D(Y)= (3,4].
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Find the domain: Y= log[sin(x-3)]+ √(16-x²)
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