solve the equation for `x , 5^(1/2)+5^(1/2 + log_5 sinx) = 15^(1/2 + log_15 cosx)`

Options

A. 50
B. 100
C. 200
D. 400

Explanation

To solve the equation \( 5^{1/2} + 5^{1/2 + \log_5 \sin x} = 15^{1/2 + \log_{15} \cos x} \), we will follow these steps:

### Step 1: Rewrite the equation
The given equation can be rewritten as:
\[ 5^{1/2} + 5^{1/2} \cdot 5^{\log_5 \sin x} = 15^{1/2} \cdot 15^{\log_{15} \cos x} \]
Using the property \( a^{\log_a b} = b \), we can simplify this to:
\[ 5^{1/2} + 5^{1/2} \sin x = 15^{1/2} \cos x \]

### Step 2: Factor out common terms
Now, we can factor out \( 5^{1/2} \) from the left side:
\[ 5^{1/2} (1 + \sin x) = 15^{1/2} \cos x \]

### Step 3: Rewrite \( 15^{1/2} \)
We know that \( 15 = 3 \cdot 5 \), thus:
\[ 15^{1/2} = (3 \cdot 5)^{1/2} = 3^{1/2} \cdot 5^{1/2} \]
Substituting this back into the equation gives:
\[ 5^{1/2} (1 + \sin x) = 3^{1/2} \cdot 5^{1/2} \cos x \]

### Step 4: Cancel \( 5^{1/2} \)
Assuming \( 5^{1/2} \neq 0 \), we can divide both sides by \( 5^{1/2} \):
\[ 1 + \sin x = 3^{1/2} \cos x \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ \sqrt{3} \cos x - \sin x = 1 \]

### Step 6: Divide the equation by 2
Dividing the entire equation by 2:
\[ \frac{\sqrt{3}}{2} \cos x - \frac{1}{2} \sin x = \frac{1}{2} \]

### Step 7: Recognize the sine subtraction identity
Recognizing that \( \frac{\sqrt{3}}{2} = \sin \frac{\pi}{3} \) and \( \frac{1}{2} = \cos \frac{\pi}{3} \), we can rewrite the left side using the sine subtraction identity:
\[ \sin\left(\frac{\pi}{3} - x\right) = \frac{1}{2} \]

### Step 8: Solve for \( x \)
The equation \( \sin\left(\frac{\pi}{3} - x\right) = \frac{1}{2} \) implies:
\[ \frac{\pi}{3} - x = \frac{\pi}{6} \quad \text{or} \quad \frac{\pi}{3} - x = \frac{5\pi}{6} \]
From the first equation:
\[ x = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6} \]
From the second equation:
\[ x = \frac{\pi}{3} - \frac{5\pi}{6} = -\frac{\pi}{2} \quad (\text{not a valid solution in the range of } x) \]

Thus, the solution is:
\[ \boxed{\frac{\pi}{6}} \]