Solve 13sin(2x) + 3cos(4x) = 9: Roots In [-π/2; Π]

Dec 12, 2025 by Admin 51 views

Hey math whizzes! Ever found yourself staring at an equation and wondering just how many solutions it's hiding? Well, today we're diving deep into one such puzzle: finding the number of roots for the equation 13sin(2x) + 3cos(4x) = 9 within the specific interval of [-π/2; π]. This isn't just about crunching numbers; it's about understanding the behavior of trigonometric functions and how they intersect with a constant. We'll break down this beast step-by-step, making sure we don't miss a single solution. So, grab your calculators, sharpen your pencils, and let's get this mathematical adventure started! We're going to explore the depths of trigonometry and emerge with a clear understanding of the roots of this intriguing equation.

Unpacking the Equation: The Heart of the Matter

Alright guys, let's get down to business with our equation: 13sin(2x) + 3cos(4x) = 9. The first thing you probably notice is that we have two different arguments for our sine and cosine functions: 2x and 4x. This is a common setup in trigonometry problems, and it usually means we'll need to use some double-angle identities to simplify things and get everything in terms of a single trigonometric function and a single argument. The goal here is to transform the equation into something more manageable, ideally a polynomial in terms of sin(2x) or cos(2x). The interval we're working with, [-π/2; π], is also crucial. It defines the boundaries within which we need to find our solutions. Remember, a trigonometric equation can have infinitely many solutions, but restricting the interval helps us find a finite, countable set.

Let's think about the double-angle identity for cosine. We know that cos(4x) can be expressed in terms of cos(2x) or sin(2x). Specifically, cos(2θ) = 1 - 2sin²(θ) or cos(2θ) = 2cos²(θ) - 1 or cos(2θ) = cos²(θ) - sin²(θ). Since our equation already has a sin(2x) term, it makes sense to express cos(4x) in terms of sin(2x). Using the identity cos(2A) = 1 - 2sin²(A), we can set A = 2x. This gives us cos(4x) = cos(2 * 2x) = 1 - 2sin²(2x). Now, let's substitute this back into our original equation:

13sin(2x) + 3(1 - 2sin²(2x)) = 9

See? We've successfully transformed the equation so that it only involves sin(2x). This is a huge step forward!

The Transformation: From Trig to Algebra

Now that we've used our trusty double-angle identity, our equation looks like this: 13sin(2x) + 3 - 6sin²(2x) = 9. The next logical step is to rearrange this into a standard quadratic form. Let's move all the terms to one side to set the equation to zero:

-6sin²(2x) + 13sin(2x) + 3 - 9 = 0

-6sin²(2x) + 13sin(2x) - 6 = 0

To make it even cleaner, let's multiply the entire equation by -1 to get a positive leading coefficient:

6sin²(2x) - 13sin(2x) + 6 = 0

Now, let y = sin(2x). Our equation becomes a familiar quadratic equation: 6y² - 13y + 6 = 0. This is where the algebra skills really kick in, guys. We need to solve this quadratic equation for y.

We can use the quadratic formula, y = [-b ± sqrt(b² - 4ac)] / 2a, or we can try factoring. Let's try factoring first. We're looking for two numbers that multiply to 6 * 6 = 36 and add up to -13. After a bit of thought, we can find that -4 and -9 fit the bill: (-4) * (-9) = 36 and (-4) + (-9) = -13. So, we can rewrite the middle term:

6y² - 4y - 9y + 6 = 0

Now, we group the terms and factor by grouping:

2y(3y - 2) - 3(3y - 2) = 0

(2y - 3)(3y - 2) = 0

This gives us two possible solutions for y:

2y - 3 = 0 => 2y = 3 => y = 3/2
3y - 2 = 0 => 3y = 2 => y = 2/3

So, we have y = 3/2 or y = 2/3. Remember, y represents sin(2x).

Back to Trigonometry: Solving for x

We found that sin(2x) = 3/2 or sin(2x) = 2/3. Now, we need to recall the properties of the sine function. The range of the sine function is [-1, 1]. This means that the value of sin(θ) can never be greater than 1 or less than -1.

Therefore, the equation sin(2x) = 3/2 has no real solutions because 3/2 is greater than 1. This is a critical observation, guys!

This leaves us with only one possibility: sin(2x) = 2/3. This is a valid value for the sine function, so we expect to find solutions for x.

Now, we need to solve sin(2x) = 2/3 for x within our given interval [-π/2; π]. First, let's consider the argument 2x. If x is in the interval [-π/2; π], then 2x will be in the interval [2*(-π/2); 2*π], which simplifies to [-π; 2π]. This is our new, expanded interval for 2x.

We are looking for values of 2x in the interval [-π; 2π] such that sin(2x) = 2/3. Since 2/3 is positive, we know that 2x must lie in the quadrants where the sine function is positive. These are the first and second quadrants.

Let α be the principal value such that sin(α) = 2/3. This value α will be in the first quadrant, specifically 0 < α < π/2. You can find this value using the inverse sine function: α = arcsin(2/3). We don't necessarily need the exact numerical value of α to count the solutions, but it's good to know it exists and where it lies.

Now let's find all possible values for 2x in the interval [-π; 2π]:

First Quadrant Solution: The first solution in the interval [0, 2π] is 2x = α. Since 0 < α < π/2, this solution is within our target interval for 2x.
Second Quadrant Solution: The second solution in the interval [0, 2π] is 2x = π - α. Since 0 < α < π/2, we have π/2 < π - α < π. This is also within our target interval for 2x.
Solutions in [-π, 0]: We also need to consider the interval [-π, 0]. The sine function is negative in the third and fourth quadrants. However, we are looking for sin(2x) = 2/3, which is positive. This means there are no solutions for 2x in the interval [-π, 0] where sin(2x) is positive. If we were to consider the full cycle starting from -2π to 2π, the solutions would repeat every 2π. Let's check the values in [-π, 0) more carefully. The sine function is positive in (0, π). In the interval [-π, 0), the sine function takes negative values. Therefore, sin(2x) = 2/3 has no solutions in [-π, 0).
Solutions beyond 2π or before -π: Our interval for 2x is [-π, 2π]. We have found two solutions within [0, 2π]. Now let's consider the relationship of sin(θ) over different intervals. The general solutions for sin(θ) = k (where -1 <= k <= 1) are θ = nπ + (-1)^n * arcsin(k), where n is an integer. In our case, 2x = nπ + (-1)^n * α, where α = arcsin(2/3).

Let's test integer values of n to see which ones give 2x within [-π, 2π]:

n = 0: 2x = 0*π + (-1)^0 * α = α. Since 0 < α < π/2, this is in [-π, 2π]. (Solution 1)
n = 1: 2x = 1*π + (-1)^1 * α = π - α. Since π/2 < π - α < π, this is in [-π, 2π]. (Solution 2)
n = 2: 2x = 2*π + (-1)^2 * α = 2π + α. This is greater than 2π, so it's outside our interval.
n = -1: 2x = -1*π + (-1)^(-1) * α = -π - α. Since α > 0, -π - α is less than -π, so it's outside our interval.
n = -2: 2x = -2*π + (-1)^(-2) * α = -2π + α. This is less than -π, so it's outside our interval.

It seems we have two solutions for 2x in the interval [-π, 2π]: 2x = α and 2x = π - α.

Let's double-check the interval for 2x. It is [-π, 2π]. This interval spans three full cycles of the sine wave if we consider the endpoints. The interval [-π, 2π] is equivalent to [-180°, 360°]. The values α and π - α are in the first and second quadrants respectively, so they are between 0 and π. Both α and π - α are within [-π, 2π].

Now, let's consider if there are any other solutions in the interval [-π, 2π]. The sine function repeats every 2π. Our interval [-π, 2π] is 3π wide. The values α and π - α are the only solutions for sin(θ) = 2/3 within the interval [0, π]. Since 2/3 is positive, sin(2x) is positive only in the first and second quadrants. The general solutions for sin(θ) = k are θ = 2kπ + arcsin(k) and θ = 2kπ + π - arcsin(k).

Let α = arcsin(2/3), where 0 < α < π/2.

We need to find 2x in [-π, 2π] such that sin(2x) = 2/3.

Possible values for 2x are:

2x = α (in Quadrant 1)
2x = π - α (in Quadrant 2)
2x = 2π + α (This is greater than 2π, so outside)
2x = 2π + (π - α) = 3π - α (This is greater than 2π, so outside)
2x = -2π + α (This is less than -π, so outside)
2x = -2π + (π - α) = -π - α (This is less than -π, so outside)

Wait, let's consider the interval for 2x again: [-π, 2π]. This interval includes negative angles. The sine function is positive in the first and second quadrants. So, we need 2x to be in an interval corresponding to the first or second quadrant within [-π, 2π].

The intervals for positive sine are generally (2kπ, (2k+1)π) for integer k.

Let's break down [-π, 2π]:

[-π, 0): Sine is negative here. No solutions.
[0, π]: Sine is positive. We have 2x = α and 2x = π - α. Both are in this range and thus in [-π, 2π].
[π, 2π]: Sine is negative here. No solutions.

So, it seems there are only two solutions for 2x in the interval [-π, 2π]: α and π - α.

Final Check and Counting the Roots

We have found two valid solutions for 2x in the interval [-π, 2π]: 2x = α and 2x = π - α, where α = arcsin(2/3). Since 0 < α < π/2, both of these values for 2x are positive and less than π. This means they fall within the interval [-π, 2π].

Now, we need to find the corresponding values of x by dividing by 2:

x = α / 2
x = (π - α) / 2

We need to ensure these values of x are within our original interval [-π/2; π].

For x = α / 2: Since 0 < α < π/2, we have 0 < α/2 < π/4. This interval (0, π/4) is entirely within [-π/2, π]. So, this is a valid solution for x.
For x = (π - α) / 2: Since 0 < α < π/2, we have π/2 < π - α < π. Dividing by 2, we get π/4 < (π - α)/2 < π/2. This interval (π/4, π/2) is also entirely within [-π/2, π]. So, this is another valid solution for x.

Let's be super thorough and check the boundaries of 2x one more time. The interval for x is [-π/2, π]. This means 2x is in [-π, 2π].

We are solving sin(2x) = 2/3 for 2x in [-π, 2π]. Let α = arcsin(2/3). We know 0 < α < π/2.

Solutions for sin(θ) = positive value occur in Quadrant 1 and Quadrant 2.

In the interval [0, 2π], the solutions are α and π - α. Both are in [0, π], which is within [-π, 2π].

Now consider the interval [-π, 0). In this interval, the sine function is negative. So, sin(2x) = 2/3 has no solutions in [-π, 0).

Therefore, the only solutions for 2x in the interval [-π, 2π] are α and π - α.

These give us two distinct values for x:

x₁ = α / 2
x₂ = (π - α) / 2

Let's confirm these x values are in [-π/2, π].

Since 0 < α < π/2:

0 < x₁ < π/4. This is within [-π/2, π].
π/4 < x₂ < π/2. This is also within [-π/2, π].

We have found two distinct roots for the equation 13sin(2x) + 3cos(4x) = 9 in the interval [-π/2; π].

So, the final answer is 2. Pretty neat, right guys? It all comes down to using the right identities, understanding the ranges of trigonometric functions, and carefully examining the given interval. Keep practicing, and you'll be solving these complex problems in no time!