Solving Trigonometric Equations with Advanced Techniques

Introduction to Trigonometric Equations and Their Solutions

Trigonometric equations are a fundamental part of calculus and mathematics, dealing with functions involving sine, cosine, and other trigonometric functions. These equations often require advanced techniques to solve, such as Newton's method, cubic function transformation, and inverse trigonometric functions.

Newton's Method for Solving Trigonometric Equations

Example with Newton's Method: Given the function u cos(x), we can solve for u and then find x.

Let u cos(x). The function is given as:

fu u^3 - u^2u - 1/3.

The derivative of the function is:

f'(u) 3u^2 - 2u 1/3^22/3.

Since f'(u) is always positive for u in mathbb{R}, the function fu is strictly increasing. Thus, the equation fu 0 has only one real solution. Starting at u_0 0.5 and iterating the formula:

u_{k 1} u_k - f(u_k) / f'(u_k),

we obtain the solution u 0.442493334024442, correct to 14 decimals.

Using Bisection and Cardano's Formula

Another method to solve trigonometric equations is by using bisection and Cardano's formula. For instance, the equation cos x c 3c^3 - 3c^2c - 1 0 has a solution near c 0.4425, which gives x 63.7365°.

Transforming and Solving with Cardano's Formula

To solve the equation ft t^3 - t^2t 1/3, we transform the unknown using t u^1/3. This results in:

u^3 - u^2 - 1/3u - 2/27 0.

Using Cardano's formula, where p 2/3 and q -2/27, we find:

u sqrt[3]{-q/2 pm sqrt{(p^3/27) - (q^2/4)}};

Thus, u 1/3sqrt[3]{4} - sqrt[3]{2}, and cos(x) u^1/3 1/3(1 - sqrt[3]{2}sqrt[3]{4}).

Advanced Geometric and Algebraic Methods

For the equation cosx - cos^2x - cos^3x 1/3, we first rearrange it as:

-1/3 - cosx - cos^2x - cos^3x 0.

Substituting y cosx - 1/3, the equation transforms into a cubic equation in y as follows:

y^3 - 2y/3 - 2/27 0.

By substituting y z - 2/9z and solving the resulting quadratic equation, we find the solutions for y and then for z, and finally returning to cosx.

The solutions are:

x pm cos^{-1}(1/3(2^{1/3}2^{1/3} - 1)) 2πn_1, n_1 in mathbb{Z},

x 2πn_2 - cos^{-1}(1/3(2^{1/3}2^{1/3} - 1)) 2πn_2, n_2 in mathbb{Z},

x pm cos^{-1}(1/3(-2^{1/3} - 1^{1/3}2^{1/3})) 2πn_3, n_3 in mathbb{Z},

x 2πn_4 - cos^{-1}(1/3(-2^{1/3} - 1^{1/3}2^{1/3})) 2πn_4, n_4 in mathbb{Z},

x pm cos^{-1}(1/3(-2^{1/3} - 2^{2/3})) 2πn_5, n_5 in mathbb{Z},

x 2πn_6 - cos^{-1}(1/3(-2^{1/3} - 2^{2/3})) 2πn_6, n_6 in mathbb{Z}.

The above solutions cover all possible values of x that satisfy the given trigonometric equation, demonstrating the power of trigonometric transformations and advanced algebraic methods.