Consider the real-valued function satisfying `2f(sinx)+f(cosx)=xdot` then the domain of `f(x)i sR` domain of `f(x)i s[-1,1]` range of `f(x)` is `[-(2pi)/3,pi/3]` range of `f(x)i sR`

Options

A. domain of `f(x)` is R
B. domain of `f(x)" is " [-1,1]`
C. range of `f(x) " is " [-(2pi)/(3),(pi)/(3)]`
D. range of `f(x)` is R

Explanation

To solve the problem, we need to analyze the function given by the equation:

\[ 2f(\sin x) + f(\cos x) = x \]

### Step 1: Substitute \( x \) with \( \frac{\pi}{2} - x \)

We start by substituting \( x \) in the original equation with \( \frac{\pi}{2} - x \):

\[ 2f(\sin(\frac{\pi}{2} - x)) + f(\cos(\frac{\pi}{2} - x)) = \frac{\pi}{2} - x \]

Using the identities \( \sin(\frac{\pi}{2} - x) = \cos x \) and \( \cos(\frac{\pi}{2} - x) = \sin x \), we can rewrite the equation as:

\[ 2f(\cos x) + f(\sin x) = \frac{\pi}{2} - x \]

### Step 2: Set up the system of equations

Now we have two equations:

1. \( 2f(\sin x) + f(\cos x) = x \) (Equation 1)
2. \( 2f(\cos x) + f(\sin x) = \frac{\pi}{2} - x \) (Equation 2)

### Step 3: Eliminate \( f(\cos x) \)

To eliminate \( f(\cos x) \), we can express \( f(\cos x) \) from Equation 1:

\[ f(\cos x) = x - 2f(\sin x) \]

Now substitute this expression into Equation 2:

\[ 2(x - 2f(\sin x)) + f(\sin x) = \frac{\pi}{2} - x \]

### Step 4: Simplify the equation

Expanding and simplifying gives:

\[ 2x - 4f(\sin x) + f(\sin x) = \frac{\pi}{2} - x \]

Combining like terms:

\[ 2x - 3f(\sin x) = \frac{\pi}{2} - x \]

### Step 5: Solve for \( f(\sin x) \)

Rearranging the equation to isolate \( f(\sin x) \):

\[ 3f(\sin x) = 3x - \frac{\pi}{2} \]

Thus, we find:

\[ f(\sin x) = x - \frac{\pi}{6} \]

### Step 6: Find the domain of \( f(x) \)

Since \( f(\sin x) \) is defined for \( \sin x \) values, the domain of \( f(x) \) is limited to the range of the sine function:

\[ \text{Domain of } f(x) = [-1, 1] \]

### Step 7: Find the range of \( f(x) \)

To find the range of \( f(x) \), we analyze the expression:

\[ f(x) = \sin^{-1}(x) - \frac{\pi}{6} \]

The range of \( \sin^{-1}(x) \) is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \). Therefore, we subtract \( \frac{\pi}{6} \) from both ends:

1. Lower bound:
\[ -\frac{\pi}{2} - \frac{\pi}{6} = -\frac{3\pi}{6} - \frac{\pi}{6} = -\frac{4\pi}{6} = -\frac{2\pi}{3} \]

2. Upper bound:
\[ \frac{\pi}{2} - \frac{\pi}{6} = \frac{3\pi}{6} - \frac{\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3} \]

Thus, the range of \( f(x) \) is:

\[ \text{Range of } f(x) = \left[-\frac{2\pi}{3}, \frac{\pi}{3}\right] \]

### Conclusion

Based on our analysis, we conclude that:

- The domain of \( f(x) \) is \( [-1, 1] \).
- The range of \( f(x) \) is \( \left[-\frac{2\pi}{3}, \frac{\pi}{3}\right] \).