Question

find the domain of log (sin^2 x)​

Answer 1

Answer:

Domain of \mathbf{f(x)=\log (\sin x)}f(x)=log(sinx) is \mathbf{\left ( 2n\pi ,(2n+1)\pi\right)}(2nπ,(2n+1)π)

Step-by-step explanation:

1. First see domain of \mathbf{f(x)=\log (\sin x)}f(x)=log(sinx) ...1)

We know that domain of ㏒a

a>0

2. Domain of equation 1) is

\mathbf{\sin x>0}sinx>0 ...2)

3. equation 2) is only varied for x value,which are

\mathbf{...(-4\pi ,-3\pi ),(-2\pi ,-\pi ),(0,\pi ),(2\pi ,3\pi ),(4\pi ,5\pi ),... }...(−4π,−3π),(−2π,−π),(0,π),(2π,3π),(4π,5π),...

4. Generalise value of x are

\mathbf{2n\pi

where n belong to integer.

5. Means domain of function are

\mathbf{x\epsilon \left ( 2n\pi ,(2n+1)\pi\right)}xϵ(2nπ,(2n+1)π)

Answer 2

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