/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 52 \(8 \sin ^{6} x+3 \cos 2 x+2 \co... [FREE SOLUTION] | 91影视

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Chapter 12: Problem 52

\(8 \sin ^{6} x+3 \cos 2 x+2 \cos 4 x+1=0\)

Short Answer

Expert verified
The detailed solution to this problem will depend on the roots of the cubic equation found in Step 3. Without specific numerical solutions for the roots of this cubic equation, it is difficult to provide a specific short answer. However, the methodology provided will guide you to find the specific solutions.

Step by step solution

01

Substitute Trigonometric Identities

Start by substituting the identities for \(\cos 2x\) and \(\cos 4x\) into the equation. So, the given equation becomes: \(8\sin^6x + 3(1 - 2\sin^2x) + 2(1 - 2\sin^2(2x)) + 1 = 0\).
02

Simplify Equation

After substitution, simplify the equation into quadratic form. So, the simplified equation becomes: \(16\sin^6x - 6\sin^2x + 3 = 0\). This is a quadratic form of equation \(Au^2 + Bu + C = 0\), where \(u = \sin^2x\). The equation can be rewritten as: \(16u^3 - 6u + 3 = 0\).
03

Solve the Quadratic Equation

With the equation in quadratic form, it can now be solved using the usual methods for solving quadratic equations, such as factoring, completing the square, or using the quadratic formula. In most cases, a numerical method such as the Newton-Raphson method may be needed to find the roots of the equation.
04

Back-substitution

The last step is to substitute back the original variable \(x\) into the solution for \(u\) to get the solutions for \(x\). Be careful to consider all potential solutions within the given domain when undoing the square root.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometric Identities
Trigonometric identities are mathematical tools used to simplify and solve trigonometric equations. In the original exercise, identities for \(\cos 2x\) and \(\cos 4x\) are utilized. These identities express cosine functions in terms of sine.
  • \(\cos 2x = 1 - 2\sin^2x\)
  • \(\cos 4x = 1 - 2\sin^2(2x)\)
These equations help rewrite complex trigonometric expressions in simpler forms. By substituting these identities, the initial equation is transformed into a simpler form that is more manageable to work with. Incorporating these identities can turn a complex equation into one that can be easily solved using algebraic methods.
Quadratic Form
The concept of quadratic form appears once an equation is simplified from its original trigonometric state. A quadratic form follows the basic structure \(Au^2 + Bu + C = 0\). In the problem provided, the equation was transformed into a quadratic form in terms of \(u = \sin^2x\):
  • \(16u^3 - 6u + 3 = 0\)
This transformation makes the equation ready for solving by polynomial-solving techniques. Quadratic forms are essential because they can often be solved by factorization, using the quadratic formula, or through other methods such as numerical approximations.
Numerical Methods
Numerical methods involve using algorithms to approximate the solutions of equations that may be otherwise difficult to solve analytically. This problem, once simplified to a polynomial form, can be tackled using such methods.
  • These methods are valuable when equations do not factor easily or when exact solutions are difficult to derive.
  • They are also used in situations where equations may have multiple solutions, making them impractical to solve by standard algebraic methods alone.
By employing numerical methods, we approximate the roots of the equation, aiming for solutions that are close enough to serve practical purposes.
Newton-Raphson Method
The Newton-Raphson method is a popular numerical technique for finding successively better approximations to the roots of a real-valued function. This method is particularly used when other solving methods are insufficient for a given equation.
  • Start with an initial guess for a root \(x_0\).
  • Iteratively apply the formula: \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\)
  • This formula calculates a new approximation from a previous one by using both the value of the function \(f(x)\) and its derivative \(f'(x)\) at \(x_n\).
This method converges quickly to a solution when the initial guess is close to the actual root, making it a powerful tool in solving equations like the one in the exercise.

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