The range of the function `sin^2x-5sinx -6` is

Answer

To find the range of the function \( f(x) = \sin^2 x - 5 \sin x - 6 \), we can follow these steps:

### Step 1: Substitute \( y = \sin x \)
Let \( y = \sin x \). The function then becomes:
\[ f(y) = y^2 - 5y - 6 \]

### Step 2: Factor the quadratic equation
Next, we can factor the quadratic equation:
\[ f(y) = (y - 6)(y + 1) \]

### Step 3: Find the critical points
To find the critical points, we can set \( f(y) = 0 \):
\[ (y - 6)(y + 1) = 0 \]
This gives us the solutions:
\[ y = 6 \quad \text{and} \quad y = -1 \]

### Step 4: Determine the range of \( y \)
Since \( y = \sin x \), we know that \( y \) can only take values in the interval \([-1, 1]\).

### Step 5: Evaluate the function at the endpoints
Now we evaluate the function at the endpoints of the interval:
1. For \( y = -1 \):
\[ f(-1) = (-1)^2 - 5(-1) - 6 = 1 + 5 - 6 = 0 \]
2. For \( y = 1 \):
\[ f(1) = (1)^2 - 5(1) - 6 = 1 - 5 - 6 = -10 \]

### Step 6: Determine the range of \( f(y) \)
The function \( f(y) \) is a quadratic that opens upwards (since the coefficient of \( y^2 \) is positive). The minimum value occurs at \( y = 1 \) and the maximum value occurs at \( y = -1 \).

Thus, the range of \( f(y) \) is:
\[ [-10, 0] \]

### Final Answer
The range of the function \( \sin^2 x - 5 \sin x - 6 \) is:
\[ [-10, 0] \]

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