If `3sinx+4cosx=2`, then find the value of `3cosx-4sinx`.

Options

A. `sqrt(21)`
B. 5
C. 0
D. `sqrt(29)`

Answer

To solve the problem where we need to find the value of \(3\cos x - 4\sin x\) given that \(3\sin x + 4\cos x = 2\), we can follow these steps:

### Step 1: Define the equations
Let:
1. \( A = 3\sin x + 4\cos x = 2 \) (given)
2. \( B = 3\cos x - 4\sin x \) (to find)

### Step 2: Square both equations
We will square both equations to eliminate the sine and cosine terms.

\[ A^2 = (3\sin x + 4\cos x)^2 = 2^2 = 4 \]

Expanding \(A^2\):
\[ (3\sin x + 4\cos x)^2 = 9\sin^2 x + 24\sin x \cos x + 16\cos^2 x \]
Thus, we have:
\[ 9\sin^2 x + 16\cos^2 x + 24\sin x \cos x = 4 \tag{1} \]

Now, for \(B\):
\[ B^2 = (3\cos x - 4\sin x)^2 \]

Expanding \(B^2\):
\[ (3\cos x - 4\sin x)^2 = 9\cos^2 x - 24\sin x \cos x + 16\sin^2 x \]
Thus, we have:
\[ 9\cos^2 x + 16\sin^2 x - 24\sin x \cos x = B^2 \tag{2} \]

### Step 3: Add equations (1) and (2)
Now, we will add equations (1) and (2):
\[ (9\sin^2 x + 16\cos^2 x + 24\sin x \cos x) + (9\cos^2 x + 16\sin^2 x - 24\sin x \cos x) = 4 + B^2 \]

Combining like terms:
\[ (9\sin^2 x + 16\sin^2 x) + (16\cos^2 x + 9\cos^2 x) + (24\sin x \cos x - 24\sin x \cos x) = 4 + B^2 \]
\[ 25 = 4 + B^2 \]

### Step 4: Solve for \(B^2\)
Now, we can solve for \(B^2\):
\[ B^2 = 25 - 4 = 21 \]

### Step 5: Find \(B\)
Taking the square root of both sides:
\[ B = \sqrt{21} \quad \text{or} \quad B = -\sqrt{21} \]

Thus, the value of \(3\cos x - 4\sin x\) is:
\[ \boxed{\sqrt{21}} \text{ or } \boxed{-\sqrt{21}} \]