General solution of `sin^(2) x-5 sin x cos x -6 cos^(2)x=0` is

Options

A. (a)`x=n pi-pi//4, n in Z` only
B. (b)`npi+tan^(-1) 6, n in Z` only
C. (c)both (a) and (2)
D. (d)none of these

Explanation

To solve the equation \( \sin^2 x - 5 \sin x \cos x - 6 \cos^2 x = 0 \), we will follow these steps:

### Step 1: Rewrite the equation
The given equation is:
\[ \sin^2 x - 5 \sin x \cos x - 6 \cos^2 x = 0 \]

### Step 2: Divide by \( \cos^2 x \)
We can divide the entire equation by \( \cos^2 x \) (assuming \( \cos x \neq 0 \)):
\[ \frac{\sin^2 x}{\cos^2 x} - 5 \frac{\sin x}{\cos x} - 6 = 0 \]
This simplifies to:
\[ \tan^2 x - 5 \tan x - 6 = 0 \]

### Step 3: Let \( t = \tan x \)
Let \( t = \tan x \). The equation now becomes:
\[ t^2 - 5t - 6 = 0 \]

### Step 4: Factor the quadratic equation
We can factor the quadratic:
\[ (t - 6)(t + 1) = 0 \]

### Step 5: Solve for \( t \)
Setting each factor to zero gives us:
\[ t - 6 = 0 \quad \Rightarrow \quad t = 6 \]
\[ t + 1 = 0 \quad \Rightarrow \quad t = -1 \]

### Step 6: Find general solutions for \( x \)
Now we find the general solutions for \( x \):
1. For \( \tan x = 6 \):
\[ x = n\pi + \tan^{-1}(6), \quad n \in \mathbb{Z} \]

2. For \( \tan x = -1 \):
\[ x = n\pi - \frac{\pi}{4}, \quad n \in \mathbb{Z} \]

### Final Answer
Thus, the general solutions for the equation \( \sin^2 x - 5 \sin x \cos x - 6 \cos^2 x = 0 \) are:
\[ x = n\pi + \tan^{-1}(6) \quad \text{and} \quad x = n\pi - \frac{\pi}{4}, \quad n \in \mathbb{Z} \]