Animated Solution for Mathematics - Functions: The equation $e^{\sin x}-e^{-\sin x}-4=0$ has

The equation $e^{\sin x}-e^{-\sin x}-4=0$ appears intimidating due to the trigonometric function $\sin x$ being trapped within an exponential. However, in JEE Advanced mathematics, such problems are often simple algebraic structures in disguise.

We observe the term $e^{\sin x}$ appearing repeatedly. Let us define a new variable, $t=e^{\sin x}$.

Because $e$ raised to any real power is always positive, we must immediately note the constraint $t>0$.

Now, consider the second term: $e^{-\sin x}$. By the laws of exponents, this is simply $\frac{1}{e^{\sin x}}$, which is $\frac{1}{t}$.

Our equation now transforms into:

Suddenly, the transcendental terror has vanished, replaced by a clean, rational algebraic equation.

To solve $t-\frac{1}{t}-4=0$, we multiply the entire equation by $t$ (which is valid since $t eq 0$). This yields:

This is a classic quadratic equation. Using the quadratic formula $t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, with $a=1,b=-4,c=-1$, we find:

We have two potential values for $t$: $t_1=2+\sqrt{5}$ and $t_2=2-\sqrt{5}$.

We must now apply our constraint $t>0$. We know $\sqrt{5}\approx 2.236$.

Thus, $t_1=2+2.236=4.236$, which is positive. However, $t_2=2-2.236=-0.236$, which is negative.

We must reject $t_2$. We are left with the condition $e^{\sin x}=2+\sqrt{5}$.

Can $e^{\sin x}$ ever reach $2+\sqrt{5}\approx 4.236$? Let us examine the range of the function $f(x)=e^{\sin x}$.

We know that for any real $x$, $-1\le \sin x\le 1$. Since the exponential function $e^u$ is strictly increasing, we apply it to the inequality:

This means $e^{\sin x}$ is bounded between $\frac{1}{e}\approx 0.368$ and $e\approx 2.718$.

Our required value, $4.236$, is far greater than the maximum possible value of $2.718$. The line $y=2+\sqrt{5}$ sits well above the peak of our exponential curve.

Therefore, there is no real $x$ that can satisfy this equation. We have proven that there are no real roots.