The value of x ` , 0 le x le (pi)/2` which satisfy the equation ` 81^( sin^(2)x)+81^(cos^(2)x)=30 ` are

Options

A. `(pi)/(3)`
B. `(pi)/(4)`
C. `(pi)/(6)`
D. `(7pi)/(18)`

Answer

To solve the equation \( 81^{\sin^2 x} + 81^{\cos^2 x} = 30 \) for \( x \) in the interval \( [0, \frac{\pi}{2}] \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ 81^{\sin^2 x} + 81^{\cos^2 x} = 30 \]
Using the identity \( \sin^2 x + \cos^2 x = 1 \), we can rewrite \( 81^{\cos^2 x} \) as \( 81^{1 - \sin^2 x} \):
\[ 81^{\sin^2 x} + 81^{1 - \sin^2 x} = 30 \]

### Step 2: Substitute \( t = 81^{\sin^2 x} \)
Let \( t = 81^{\sin^2 x} \). Then, \( 81^{\cos^2 x} = \frac{81}{t} \) (since \( 81^{\cos^2 x} = 81^{1 - \sin^2 x} = \frac{81^{\sin^2 x}}{81^{\sin^2 x}} \cdot 81 \)).
Substituting this into the equation gives:
\[ t + \frac{81}{t} = 30 \]

### Step 3: Multiply through by \( t \)
To eliminate the fraction, multiply through by \( t \):
\[ t^2 + 81 = 30t \]
Rearranging this gives:
\[ t^2 - 30t + 81 = 0 \]

### Step 4: Solve the quadratic equation
Now we can solve the quadratic equation \( t^2 - 30t + 81 = 0 \) using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -30, c = 81 \):
\[ t = \frac{30 \pm \sqrt{(-30)^2 - 4 \cdot 1 \cdot 81}}{2 \cdot 1} \]
Calculating the discriminant:
\[ t = \frac{30 \pm \sqrt{900 - 324}}{2} = \frac{30 \pm \sqrt{576}}{2} = \frac{30 \pm 24}{2} \]
This gives us two solutions:
\[ t_1 = \frac{54}{2} = 27 \quad \text{and} \quad t_2 = \frac{6}{2} = 3 \]

### Step 5: Relate \( t \) back to \( x \)
Recall that \( t = 81^{\sin^2 x} \). We have:
1. \( 81^{\sin^2 x} = 27 \)
2. \( 81^{\sin^2 x} = 3 \)

#### For \( t_1 = 27 \):
\[ 81^{\sin^2 x} = 3^3 \implies 3^{4\sin^2 x} = 3^3 \implies 4\sin^2 x = 3 \implies \sin^2 x = \frac{3}{4} \implies \sin x = \frac{\sqrt{3}}{2} \]
Thus, \( x = \frac{\pi}{3} \).

#### For \( t_2 = 3 \):
\[ 81^{\sin^2 x} = 3^1 \implies 3^{4\sin^2 x} = 3^1 \implies 4\sin^2 x = 1 \implies \sin^2 x = \frac{1}{4} \implies \sin x = \frac{1}{2} \]
Thus, \( x = \frac{\pi}{6} \).

### Final Step: Conclusion
The values of \( x \) that satisfy the original equation in the interval \( [0, \frac{\pi}{2}] \) are:
\[ x = \frac{\pi}{3} \quad \text{and} \quad x = \frac{\pi}{6} \]