The number of integral values of k for which the equation 7 cos x + 5 sin x = 2k + 1 has a solution

Q: Wait, why can't we just solve for $x$ directly?

That is a great question! If you try to solve for $x$ directly, you will find yourself dealing with transcendental equations that are not easily solvable using standard algebraic methods. In JEE problems like this, the goal is to use the properties of the function's range to find the condition for existence, rather than finding the specific value of $x$.

Q: How do I know if the range is inclusive or exclusive?

The function $f(x) = a \cos x + b \sin x$ is a continuous, periodic function. It oscillates smoothly between its maximum and minimum values. Because it is continuous, it must take on every value within its range, including the endpoints. Therefore, the range is always inclusive: $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.

Q: Do I need to worry about the period of the function?

Not at all! The range of a trigonometric function depends only on its amplitude, not its period. Whether the wave oscillates quickly or slowly, its maximum and minimum heights remain the same. You can safely ignore the period for this specific problem.

Question

The number of integral values of $k$ for which the equation $7 \cos x + 5 \sin x = 2k + 1$ has a solution is

Options:

(A) $4$

(B) $8$

(C) $10$

(D) $12$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

My dear student, let us embark on a journey to decode this trigonometric puzzle. At first glance, the equation $7 \cos x + 5 \sin x = 2k + 1$ might seem daunting, but it is actually a beautiful dance between a wave and a line.

On the left-hand side, we have $7 \cos x + 5 \sin x$. This is not just a random collection of terms; it is a single, continuous trigonometric wave.

On the right-hand side, we have $2k + 1$, which, for any fixed integer $k$, is simply a constant horizontal line. The problem asks us to find the number of integral values of $k$ for which this line intersects our wave.

If the line is too high or too low, it will never touch the wave, and there will be no solution. If it lies within the wave's reach, it will intersect it, and a solution will exist.

### The Boundary Condition

To find where the line can exist, we must determine the range of our wave. We use the fundamental property that any expression of the form $a \cos x + b \sin x$ oscillates between $-\sqrt{a^2 + b^2}$ and $\sqrt{a^2 + b^2}$.

In our case, $a = 7$ and $b = 5$. Let us calculate the amplitude:

$$ \sqrt{7^2 + 5^2} = \sqrt{49 + 25} = \sqrt{74} $$

Thus, our wave is strictly bounded between $-\sqrt{74}$ and $\sqrt{74}$. To make this tangible, let us approximate $\sqrt{74}$. Since $8^2 = 64$ and $9^2 = 81$, we know $\sqrt{74}$ is approximately $8.6$. So, our wave lives in the interval $[-8.6, 8.6]$.

### The Algebraic Squeeze

Now, for the equation to have a solution, the horizontal line $y = 2k + 1$ must fall within this interval. This gives us the inequality:

$$ -\sqrt{74} \le 2k + 1 \le \sqrt{74} $$

Substituting our approximation, we get $-8.6 \le 2k + 1 \le 8.6$. Our goal is to isolate $k$.

First, we subtract $1$ from all parts of the inequality:

$$ -9.6 \le 2k \le 7.6 $$

Next, we divide by $2$:

$$ -4.8 \le k \le 3.8 $$

This is the golden range for $k$. Any integer $k$ that sits within this interval will guarantee that our line intersects the wave.

### The Integer Hunt

Now, we simply count the integers in the interval $[-4.8, 3.8]$. Let us list them carefully: $-4, -3, -2, -1, 0, 1, 2, 3$.

If we count them, we find there are $4$ negative integers, $1$ zero, and $3$ positive integers. Adding these together, $4 + 1 + 3 = 8$.

There are exactly **$8$ integral values of $k$** for which the equation has a solution. You see, my friend, when you break down a complex problem into its geometric and algebraic components, the path to the solution becomes clear and elegant.

Animated Solution for Mathematics - Trigonometry: The number of integral values of $k$ for which the equation $7 cos x + 5 sin x = 2 k + 1$ has a solution is

Select Answer:

Visualized Solution

Analyze the Equation Structure

Recall the Range Formula

Calculate the Bounds

Approximate the Square Root

Set up the Intersection Condition

Formulate the Inequality

Isolate the Term $2 k$

Solve for $k$

Visualize the Valid Range

Identify Integral Values

Final Count and Conclusion

The Sigma Insight: Trigonometric Ratios and Identities

Analyzing the Setup

The Boundary Condition

The Algebraic Squeeze

The Integer Hunt

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Animated Solution for Mathematics - Trigonometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of integral values of k for which the equation 7cosx+5sinx=2k+1 has a solution is