Animated Solution for Mathematics - Trigonometry: The sum of all values of $x$ in $[0,2\pi ]$, for which $\sin x+\sin 2x+\sin 3x+\sin 4x=0$, is equal to :

Solving Trigonometric Sums

Imagine standing before a complex trigonometric equation:

At first glance, it looks like a chaotic mess of oscillating waves. You might be tempted to expand $\sin 2x$ or $\sin 3x$ using multiple-angle formulas, but stop! That path leads to a polynomial nightmare.

In JEE Advanced, the beauty of trigonometry lies not in brute force, but in recognizing the hidden rhythm of the angles.

The Art of Strategic Grouping

Look at the angles: $x,2x,3x,$ and $4x$. There is a beautiful arithmetic progression here.

If we pair the first term with the last, $x+4x=5x$. If we pair the middle two, $2x+3x=5x$. This is our golden ticket.

By grouping them as $(\sin x+\sin 4x)+(\sin 2x+\sin 3x)=0$, we are setting the stage for a common factor to emerge. We are not just rearranging terms; we are orchestrating a mathematical harmony.

The Sum-to-Product Transformation

Now, we invoke the powerful sum-to-product identity:

Applying this to our pairs, the first group becomes $2\sin \frac{5x}{2}\cos \frac{3x}{2}$, and the second becomes $2\sin \frac{5x}{2}\cos \frac{x}{2}$. Suddenly, the chaos vanishes.

We have a common factor of $2\sin \frac{5x}{2}$ staring us in the face. Factoring this out, we get:

The equation is now a product, which is the most desirable state for any equation.

The Final Factoring

We are not done yet. Inside the bracket, we have $\cos \frac{3x}{2}+\cos \frac{x}{2}$. Applying the sum-to-product identity for cosines, $\cos C+\cos D=2\cos \frac{C+D}{2}\cos \frac{C-D}{2}$, we transform the bracket into $2\cos x\cos \frac{x}{2}$.

Our equation is now fully factored:

This is the elegance of trigonometry. We have reduced a sum of four terms into a product of three simple factors.

Hunting the Roots

Now, we solve for each factor independently. First, $\sin \frac{5x}{2}=0$ implies $\frac{5x}{2}=n\pi$, or $x=\frac{2n\pi }{5}$.

For $n=0,1,2,3,4,5$, we get $x=0,\frac{2\pi }{5},\frac{4\pi }{5},\frac{6\pi }{5},\frac{8\pi }{5},2\pi$. That is six solutions.

Second, $\cos x=0$ gives $x=\frac{\pi }{2},\frac{3\pi }{2}$. Finally, $\cos \frac{x}{2}=0$ gives $x=\pi$.

Summing these up: the first set sums to $6\pi$, the second to $2\pi$, and the third to $\pi$. The grand total is $9\pi$.

You have navigated the complexity and emerged with a clean, elegant result. Remember, in JEE, always look for the symmetry before you reach for the expansion formulas.