The number of solutions of the equation \(32^{\tan^2 x} + 32^{\sec^2 x} = 81\), \(0 \le x \le \frac{\pi}{4}\) is :
Show Hint
In transcendental equations, instead of solving for the exact value of \(x\), check if the required range of the intermediate variable (\(\tan^2 x\)) overlaps with the range of the given numeric value (\(\log_{32} \frac{27}{11}\)).
Step 1: Understanding the Concept:
We use trigonometric identities to simplify exponents to a single variable. Then we solve the resulting algebraic equation within the given restricted domain of \(x\). Step 2: Key Formula or Approach:
Identify the relationship: \(\sec^2 x = 1 + \tan^2 x\).
Let \(32^{\tan^2 x} = t\). Then \(32^{\sec^2 x} = 32^{1 + \tan^2 x} = 32^1 \cdot 32^{\tan^2 x} = 32t\). Step 3: Detailed Explanation:
Substituting \(t\) into the equation:
\[ t + 32t = 81 \]
\[ 33t = 81 \implies t = \frac{81}{33} = \frac{27}{11} \]
Now, equate back to the expression in \(x\):
\[ 32^{\tan^2 x} = \frac{27}{11} \]
Taking \(\log_{32}\) on both sides:
\[ \tan^2 x = \log_{32} \left(\frac{27}{11}\right) \]
We need to check if this has solutions for \(0 \le x \le \frac{\pi}{4}\).
For \(x \in [0, \frac{\pi}{4}]\), \(\tan x \in [0, 1]\), so \(\tan^2 x \in [0, 1]\).
Note that \(1<\frac{27}{11}<32\).
Since \(32^0<\frac{27}{11}<32^1\), the value of \(\log_{32} (\frac{27}{11})\) lies between \(0\) and \(1\).
Since \(\tan^2 x\) takes a value in the interval \((0, 1)\), there is exactly one value of \(\tan^2 x\) that satisfies this.
In the domain \([0, \frac{\pi}{4}]\), \(\tan x\) is non-negative, so \(\tan x = \sqrt{\log_{32} (\frac{27}{11})}\) gives exactly one unique solution for \(x\). Step 4: Final Answer:
The number of solutions is 1.
