Animated Solution for Mathematics - Trigonometry: The number of all possible triplets $(a_1,a_2,a_3)$ such that $a_1+a_2\cos (2x)+a_3{\sin }^2(x)=0$ for all $x$ is

Imagine you are standing before a mathematical puzzle that seems to defy simple logic. You are given the equation $a_1+a_2\cos (2x)+a_3{\sin }^2(x)=0$ and told it must hold for all $x$.

At first glance, your instinct might be to solve for $x$. But stop! The phrase "for all $x$" is the most powerful sentence in this problem.

It transforms the equation from a simple puzzle into a profound statement of identity. We are not looking for a specific $x$; we are looking for the hidden structure that makes this expression vanish regardless of the input.

To solve this, we must first recognize that our terms are speaking different languages. We have a $\cos (2x)$ and a ${\sin }^2(x)$.

They are like two people trying to communicate in different dialects. To make sense of the identity, we must bring them onto common ground.

This is where the double-angle identity for cosine comes to our rescue:

This formula is the bridge that connects our two terms. By substituting this into our original equation, we replace the cosine term with something that speaks the language of ${\sin }^2(x)$.

Let's perform the substitution:

Now, we distribute the $a_2$ term. It is crucial to be careful with the signs here—a small slip could derail the entire process. We get:

Now, let's group the terms. We have the constant part, $(a_1+a_2)$, and the part dependent on ${\sin }^2(x)$, which is $(a_3-2a_2){\sin }^2(x)$. Our equation now looks like this:

Here is the core of the problem. We have an expression of the form $A+B{\sin }^2(x)=0$. For this to be true for every single value of $x$, the coefficients $A$ and $B$ must independently be zero.

Why? Because $1$ and ${\sin }^2(x)$ are linearly independent functions. They do not "cancel each other out" unless their individual coefficients are zero.

So, we set the constant part to zero:

Then, we set the coefficient of ${\sin }^2(x)$ to zero:

We have successfully expressed $a_1$ and $a_3$ in terms of $a_2$. Let's introduce a parameter $k$ such that $a_2=k$. Then, our triplet $(a_1,a_2,a_3)$ becomes $(-k,k,2k)$.

Think about what this means. For any real number $k$ you choose—be it $1,0,-5,$ or even $\pi$—you will generate a valid triplet that satisfies the original equation.

Because $k$ can be any real number, there are infinitely many such triplets. This is the elegance of the identity: it doesn't restrict us to a single answer; it reveals an entire family of solutions, all bound by the same beautiful mathematical law.