Animated Solution for Mathematics - Trigonometry: The number of solutions, of the equation $e^{\sin x}-2e^{-\sin x}=2$ is

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to dissect an equation that, at first glance, seems like a standard algebraic problem but is actually a masterclass in understanding domains and ranges.

The equation is $e^{\sin x}-2e^{-\sin x}=2$. It looks intimidating, but remember, in mathematics, complexity is often just a mask for a simpler structure waiting to be revealed.

When you see an expression like $e^{\sin x}$ repeating, your first instinct should be to simplify. Let us define a new variable, $t$, such that $t=e^{\sin x}$.

This transforms our equation into $t-2e^{-\sin x}=2$. Since $e^{-\sin x}$ is the reciprocal of $e^{\sin x}$, we can write this as:

Suddenly, the transcendental nature of the equation vanishes, and we are left with a clean, algebraic relationship.

However, we must respect the boundaries of our new variable. The sine function, $\sin x$, is restricted to the interval $[-1,1]$.

Because $t=e^{\sin x}$, the range of $t$ is constrained by the exponential function over this interval. The minimum value is $e^{-1}=\frac{1}{e}$ and the maximum value is $e^1=e$.

Thus, our valid zone for $t$ is $[\frac{1}{e},e]$. Any solution we find for $t$ that falls outside this interval must be discarded.

With our domain secured, let us return to the equation $t-\frac{2}{t}=2$. To clear the fraction, we multiply the entire equation by $t$, yielding:

Rearranging this into the standard quadratic form, we get $t^2-2t-2=0$. To find its roots, we deploy the quadratic formula:

Substituting $a=1$, $b=-2$, and $c=-2$, we calculate the discriminant: $b^2-4ac=(-2{)}^2-4(1)(-2)=4+8=12$.

Thus, the roots are:

Now, we must test our candidates. First, consider $t_1=1-\sqrt{3}$.

Since $\sqrt{3}\approx 1.732$, $t_1\approx -0.732$. As $t=e^{\sin x}$ must be positive, $t_1$ is impossible and must be rejected.

Now, consider $t_2=1+\sqrt{3}\approx 2.732$. We must check if it lies within our valid zone $[\frac{1}{e},e]$.

We know $e\approx 2.718$. Comparing the two, we see that $2.732>2.718$.

Our root $t_2$ has crossed the boundary; it is slightly larger than the maximum possible value of $e^{\sin x}$. Consequently, this root is also invalid.

Since both potential roots fail the domain test, there is no real value of $x$ that satisfies the original equation. The number of solutions is exactly 0.

This problem teaches us a vital lesson: always respect the domain. The algebra might give you answers, but the geometry of the function decides if they are real.