If n is the number of solutions of the equation 2 cos x (4 sin(π/4 + x) sin(π/4 - x) - 1) = 1, x ∈ [

Q: Wait, why do we change the domain to $3x \in [0, 3\pi]$? Isn't the domain just $x \in [0, \pi]$?

That is a fantastic question! When we solve for $3x$, we are essentially looking for all angles in the range $[0, 3\pi]$ that satisfy the cosine condition. If we only looked at $[0, \pi]$, we would miss the solutions that exist in the second and third cycles of the cosine function. Always adjust your domain based on the argument of the trigonometric function!

Q: How did you know to convert $\sin^2 x$ to $\cos^2 x$ instead of doing something else?

Great intuition! The goal in solving trigonometric equations is to reduce them to a single ratio. We saw a $\cos x$ outside the bracket and a $\sin^2 x$ inside. Converting $\sin^2 x$ to $1 - \cos^2 x$ allows us to express the entire equation as a polynomial in $\cos x$, which is much easier to solve.

Q: Is there a faster way to find the sum of solutions without finding each one individually?

While you can sometimes use properties of the roots of polynomials, finding the individual solutions here is quite straightforward once you have the simplified equation $2 \cos 3x = 1$. For JEE Advanced, it is often safer and more reliable to find the specific values of $x$ to ensure you don't miss any or include extraneous ones.

Question

If $n$ is the number of solutions of the equation $2 \cos x (4 \sin(\frac{\pi}{4} + x) \sin(\frac{\pi}{4} - x) - 1) = 1, x \in [0, \pi]$ and $S$ is the sum of all these solutions, then the ordered pair $(n, S)$ is :

Options:

(A) $(3, 13\pi/9)$

(B) $(2, 2\pi/3)$

(C) $(2, 8\pi/9)$

(D) $(3, 5\pi/3)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex trigonometric equation:

$$2 \cos x \left(4 \sin\left(\frac{\pi}{4} + x\right) \sin\left(\frac{\pi}{4} - x\right) - 1\right) = 1$$

It looks intimidating, but in the world of JEE Advanced, complexity is often just a mask for elegance. Our goal is to peel back these layers.

The first thing that should catch your eye is the product of the two sine terms: $\sin(\frac{\pi}{4} + x) \sin(\frac{\pi}{4} - x)$. This is a classic structure that utilizes the identity $\sin(A+B)\sin(A-B) = \sin^2 A - \sin^2 B$. By identifying this, we have already found the key to the lock.

### The Transformation

Let us apply this identity. With $A = \frac{\pi}{4}$ and $B = x$, the expression inside the bracket becomes:

$$4\left(\sin^2 \frac{\pi}{4} - \sin^2 x\right) - 1$$

We know that $\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$, so $\sin^2(\frac{\pi}{4}) = \frac{1}{2}$. Substituting this, the bracket simplifies beautifully:

$$4\left(\frac{1}{2} - \sin^2 x\right) - 1 = 2 - 4 \sin^2 x - 1 = 1 - 4 \sin^2 x$$

Now, our equation is $2 \cos x (1 - 4 \sin^2 x) = 1$. To achieve harmony between the trigonometric functions, we use the fundamental identity $\sin^2 x = 1 - \cos^2 x$.

Substituting this, we get:

$$2 \cos x (1 - 4(1 - \cos^2 x)) = 1$$

This expands to:

$$2 \cos x (4 \cos^2 x - 3) = 1$$

### The Triple Angle Reveal

This is the moment of truth. When we distribute the $2 \cos x$, we get:

$$8 \cos^3 x - 6 \cos x = 1$$

If we factor out a $2$, we are left with:

$$2(4 \cos^3 x - 3 \cos x) = 1$$

The expression inside the bracket is the triple angle identity for cosine: $\cos 3x = 4 \cos^3 x - 3 \cos x$. The entire equation collapses into:

$$2 \cos 3x = 1 \Rightarrow \cos 3x = \frac{1}{2}$$

### The Domain Trap

Now, we must be careful. The problem states $x \in [0, \pi]$, which implies that $3x$ must lie in the interval $[0, 3\pi]$. We need to find all values of $3x$ in this range where $\cos 3x = \frac{1}{2}$.

In the first cycle $[0, 2\pi]$, cosine is positive at $3x = \frac{\pi}{3}$ and $3x = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$. In the next cycle $[2\pi, 3\pi]$, we have $3x = 2\pi + \frac{\pi}{3} = \frac{7\pi}{3}$.

Dividing these values by $3$, we obtain the solutions for $x$:

$$x = \frac{\pi}{9}, \frac{5\pi}{9}, \frac{7\pi}{9}$$

### Conclusion

We have found $n = 3$ solutions. The sum $S$ is calculated as follows:

$$S = \frac{\pi}{9} + \frac{5\pi}{9} + \frac{7\pi}{9} = \frac{13\pi}{9}$$

The final ordered pair $(n, S)$ is **$(3, \frac{13\pi}{9})$**. Remember, the path to the solution is not about brute force; it is about recognizing the patterns hidden within the math.

Animated Solution for Mathematics - Trigonometry: If $n$ is the number of solutions of the equation $2 cos x (4 sin (\frac{π}{4} + x) sin (\frac{π}{4} - x) - 1) = 1, x \in [0, π]$ and $S$ is the sum of all these solutions, then the ordered pair $(n, S)$ is :

Select Answer:

Visualized Solution

Analyze the Equation Structure

Apply Sine Product Identity

Substitute Known Values

Simplify the Inner Bracket

Convert Sine to Cosine

Expand and Group Terms

Identify the Triple Angle Identity

Solve for Cosine 3x

Find Solutions for 3x

Calculate x and Sum of Solutions

Final Result and Key Takeaway

The Sigma Insight: General Solution of Trigonometric Equations

Analyzing the Setup

The Transformation

The Triple Angle Reveal

The Domain Trap

Conclusion

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Select Answer:

Visualized Solution

Analyze the Equation Structure

Apply Sine Product Identity

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Conclusion

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