Animated Solution for Mathematics - Trigonometry: The number of solutions of the equation $32^{{\tan }^{2}x}+32^{{\sec }^{2}x}=81,0\le x\le \frac{\pi }{4}$ is:

The given equation is $32^{{\tan }^{2}x}+32^{{\sec }^{2}x}=81$, where we seek the number of solutions in the interval $x\in [0,\frac{\pi }{4}]$.

Many students attempt to solve for $x$ directly, but the key to this problem lies in identifying the relationship between the trigonometric functions in the exponents.

The fundamental Pythagorean identity provides the necessary link: ${\sec }^{2}x=1+{\tan }^{2}x$.

By substituting this identity into the original equation, we transform the expression into a single-variable problem:

Using the exponent rule $a^{m+n}=a^m\cdot a^n$, we can rewrite the second term as:

Substituting this back into our equation yields:

Let $u=32^{{\tan }^{2}x}$. The equation simplifies to:

Solving for $u$, we obtain:

We now analyze the function $f(x)=32^{{\tan }^{2}x}$ on the interval $x\in [0,\frac{\pi }{4}]$ to determine the number of solutions for $f(x)=\frac{27}{11}$.

As $x$ increases from $0$ to $\frac{\pi }{4}$, ${\tan }^{2}x$ increases monotonically from $0$ to $1$. Since the base $32>1$, the function $f(x)=32^{{\tan }^{2}x}$ is strictly increasing on this interval.

We evaluate the function at the boundaries:

The range of $f(x)$ on the given interval is $[1,32]$. Since the target value $\frac{27}{11}\approx 2.45$ lies strictly within the interval $(1,32)$, the Intermediate Value Theorem guarantees that the function attains this value.

Because the function is strictly increasing, it can take this value exactly once. Therefore, there is exactly 1 solution.