MHB Find the domain of the function .

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To find the domain of the function f(A) = (1 + sinA) / (1 - sinA) given its range is {0, 1, 3}, the equation f(A) = 0 leads to solving 1 + sinA = 0, resulting in sinA = -1. For f(A) = 1, the equation simplifies to 1 + sinA = 1 - sinA, giving sinA = 0. Lastly, for f(A) = 3, the equation 1 + sinA = 3(1 - sinA) leads to sinA = 2/4 or sinA = 1/2. The domain is thus determined by the values of A that satisfy these conditions. The analysis reveals the specific angles corresponding to these sine values, defining the complete domain of the function.
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1) Given that the range of function f(A) = \frac{1 + sinA}{1 - sinA} is { 0,1,3} . Find the domain of the function .
 
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Riwaj said:
1) Given that the range of function f(A) = \frac{1 + sinA}{1 - sinA} is { 0,1,3} . Find the domain of the function .

Let f(A) = 0 and solve for A.

Do the same for 1 and 3.
 
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Thread 'Erroneously finding discrepancy in transpose rule'

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