`81^(cos^2x)+81^(sin^2x)=30` . then no. of solutions in `x in (o,pi)`

Options

A. 0
B. 2
C. 8
D. 4

Answer

To solve the equation \( 81^{\cos^2 x} + 81^{\sin^2 x} = 30 \) and find the number of solutions for \( x \) in the interval \( (0, \pi) \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ 81^{\cos^2 x} + 81^{\sin^2 x} = 30 \]
Since \( 81 = 3^4 \), we can rewrite the equation as:
\[ (3^4)^{\cos^2 x} + (3^4)^{\sin^2 x} = 30 \]
This simplifies to:
\[ 3^{4\cos^2 x} + 3^{4\sin^2 x} = 30 \]

### Step 2: Introduce a substitution
Let \( t = 3^{4\sin^2 x} \). Then, we have:
\[ 3^{4\cos^2 x} = \frac{81}{t} \]
Substituting this into our equation gives:
\[ \frac{81}{t} + t = 30 \]

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

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

### Step 5: Back substitute for \( \sin^2 x \)
Recall that \( t = 3^{4\sin^2 x} \). Thus, we have two cases:
1. \( 3^{4\sin^2 x} = 27 \)
2. \( 3^{4\sin^2 x} = 3 \)

#### Case 1: \( 3^{4\sin^2 x} = 27 \)
Taking logarithm base 3:
\[ 4\sin^2 x = 3 \implies \sin^2 x = \frac{3}{4} \implies \sin x = \pm \frac{\sqrt{3}}{2} \]
The solutions for \( x \) in \( (0, \pi) \) are:
- \( x = \frac{\pi}{3} \) (for \( \sin x = \frac{\sqrt{3}}{2} \))
- \( x = \frac{2\pi}{3} \) (for \( \sin x = -\frac{\sqrt{3}}{2} \) is not valid in this interval)

#### Case 2: \( 3^{4\sin^2 x} = 3 \)
Taking logarithm base 3:
\[ 4\sin^2 x = 1 \implies \sin^2 x = \frac{1}{4} \implies \sin x = \pm \frac{1}{2} \]
The solutions for \( x \) in \( (0, \pi) \) are:
- \( x = \frac{\pi}{6} \) (for \( \sin x = \frac{1}{2} \))
- \( x = \frac{5\pi}{6} \) (for \( \sin x = -\frac{1}{2} \) is not valid in this interval)

### Step 6: Count the valid solutions
From both cases, the valid solutions in the interval \( (0, \pi) \) are:
- \( x = \frac{\pi}{3} \)
- \( x = \frac{\pi}{6} \)
- \( x = \frac{5\pi}{6} \)

Thus, we have a total of **three valid solutions**.

### Final Answer:
The number of solutions in \( x \) in the interval \( (0, \pi) \) is **3**.

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