Animated Solution for Mathematics - Trigonometry: The solution set of the system of equations $x+y=\frac{2\pi }{3}$, $\cos x+\cos y=\frac{3}{2}$, where $x$ and $y$ are real, is ..........

Imagine you are standing at the base of a mountain, staring up at a path that seems to twist and turn into a complex system of equations. You are given the following constraints:

At first glance, it looks like a standard algebra problem, but there is a hidden trap waiting for the unwary. Let us embark on this journey together to uncover the truth.

When we face a system involving the sum of trigonometric functions, our first instinct should be to look for a bridge. That bridge is the sum-to-product identity:

This identity is powerful because it transforms a sum into a product, allowing us to isolate the variables in a way that makes the underlying structure visible. By applying this to our second equation, we obtain:

Now, look at the first equation: $x+y=\frac{2\pi }{3}$. This is not just a constraint; it is a gift. By substituting this into our identity, the expression becomes:

Notice how the angle inside the first cosine simplifies beautifully. Since $\frac{2\pi }{3}$ divided by $2$ is simply $\frac{\pi }{3}$, and we know that $\cos (\frac{\pi }{3})=\frac{1}{2}$, we substitute this value:

The $2$ and the $\frac{1}{2}$ cancel out, leaving us with the stark, simple reality:

Here is where the thrill of the problem lies. We have arrived at the equation $\cos (\theta )=1.5$, where $\theta =\frac{x-y}{2}$.

But stop and think for a moment. What do we know about the cosine function? For any real angle $\theta$, the value of $\cos (\theta )$ is trapped in the interval $[-1,1]$.

It can never, ever exceed $1$. By demanding that $\cos (\theta )=1.5$, the system is asking for the impossible. The geometry of the cosine wave simply does not reach the height of $1.5$.

Because the cosine function is bounded, there are no real values of $x$ and $y$ that can satisfy this condition. The system is inconsistent.

In the language of mathematics, we say the solution set is the empty set, denoted by $\varphi$. This problem teaches us a vital lesson in JEE preparation: always check the domain and range of your functions before you get lost in the algebra.

Sometimes, the most elegant solution is realizing that the path you are on leads to a wall, and that wall is the answer itself. Keep exploring, keep questioning, and never let the complexity of an equation hide the simple truths of mathematics.