If `0le x le pi` and `81^(sin^2 x) + 81^(cos^2x) = 30`, then x is equal to

Options

A. `pi//6`
B. `pi//3`
C. `5pi//6`
D. `2pi//3`

Answer

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

### Step 1: Rewrite the equation
We know that \( \sin^2 x + \cos^2 x = 1 \). Therefore, we can express \( \cos^2 x \) as \( 1 - \sin^2 x \):
\[ 81^{\sin^2 x} + 81^{1 - \sin^2 x} = 30 \]

### Step 2: Simplify the equation
Let \( y = 81^{\sin^2 x} \). Then, we can rewrite the equation as:
\[ y + 81^{1 - \sin^2 x} = 30 \]
Since \( 81^{1 - \sin^2 x} = \frac{81}{y} \), we can substitute this into the equation:
\[ y + \frac{81}{y} = 30 \]

### Step 3: Multiply through by \( y \)
To eliminate the fraction, multiply the entire equation by \( y \):
\[ y^2 + 81 = 30y \]

### Step 4: Rearrange the equation
Rearranging gives us a standard quadratic equation:
\[ y^2 - 30y + 81 = 0 \]

### Step 5: Solve the quadratic equation
We can use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -30, c = 81 \):
\[ y = \frac{30 \pm \sqrt{(-30)^2 - 4 \cdot 1 \cdot 81}}{2 \cdot 1} \]
\[ y = \frac{30 \pm \sqrt{900 - 324}}{2} \]
\[ y = \frac{30 \pm \sqrt{576}}{2} \]
\[ y = \frac{30 \pm 24}{2} \]

### Step 6: Calculate the values of \( y \)
Calculating the two possible values:
1. \( y = \frac{54}{2} = 27 \)
2. \( y = \frac{6}{2} = 3 \)

### Step 7: Convert back to \( \sin^2 x \)
Recall that \( y = 81^{\sin^2 x} \):
1. For \( y = 27 \):
\[ 81^{\sin^2 x} = 27 \implies 3^{4\sin^2 x} = 3^3 \implies 4\sin^2 x = 3 \implies \sin^2 x = \frac{3}{4} \]
Thus, \( \sin x = \frac{\sqrt{3}}{2} \) or \( \sin x = -\frac{\sqrt{3}}{2} \) (but we discard the negative in the range \( 0 \leq x \leq \pi \)):
\[ x = \frac{\pi}{3} \]

2. For \( y = 3 \):
\[ 81^{\sin^2 x} = 3 \implies 3^{4\sin^2 x} = 3^1 \implies 4\sin^2 x = 1 \implies \sin^2 x = \frac{1}{4} \]
Thus, \( \sin x = \frac{1}{2} \) or \( \sin x = -\frac{1}{2} \) (again, we discard the negative):
\[ x = \frac{\pi}{6} \quad \text{or} \quad x = \frac{5\pi}{6} \]

### Step 8: List all solutions
The solutions for \( x \) in the interval \( [0, \pi] \) are:
\[ x = \frac{\pi}{3}, \quad x = \frac{\pi}{6}, \quad x = \frac{5\pi}{6} \]

### Final Answer
Thus, the values of \( x \) that satisfy the equation are:
\[ x = \frac{\pi}{3}, \quad x = \frac{\pi}{6}, \quad x = \frac{5\pi}{6} \]