Math Problem Statement

Solve the following equations: a) cos(2x - π) = cos(π/3), b) 4cos^2(x) - 3 = 0

Solution

The image contains two trigonometric equations in Finnish:

Task: "Ratkaise yhtälö."
Translation: "Solve the equation."

The given equations are:

a) cos(2xπ)=cos(π3)\cos(2x - \pi) = \cos\left(\frac{\pi}{3}\right)
b) 4cos2x3=04 \cos^2 x - 3 = 0

Let's solve these equations step by step:


a) cos(2xπ)=cos(π3)\cos(2x - \pi) = \cos\left(\frac{\pi}{3}\right)

First, recall the identity for the cosine function: cos(θ)=cos(2kπ±θ)\cos(\theta) = \cos(2k\pi \pm \theta) The cosine of an angle shifted by π\pi satisfies the following property: cos(2xπ)=cos(2x)\cos(2x - \pi) = -\cos(2x)

So, we can rewrite the equation as: cos(2x)=cos(π3)-\cos(2x) = \cos\left(\frac{\pi}{3}\right)

From trigonometry, cos(π3)=12\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}, so the equation becomes: cos(2x)=12-\cos(2x) = \frac{1}{2} cos(2x)=12\cos(2x) = -\frac{1}{2}

Now, solve for 2x2x: cos(2x)=12\cos(2x) = -\frac{1}{2}

The general solution for cos(2x)=12\cos(2x) = -\frac{1}{2} is: 2x=2kπ±2π32x = 2k\pi \pm \frac{2\pi}{3}

Thus, divide both sides by 2 to find xx: x=kπ±π3x = k\pi \pm \frac{\pi}{3}

This gives the general solution for xx: x=kπ±π3,kZx = k\pi \pm \frac{\pi}{3}, \quad k \in \mathbb{Z}


b) 4cos2x3=04 \cos^2 x - 3 = 0

Rearrange the equation: 4cos2x=34 \cos^2 x = 3

Divide by 4: cos2x=34\cos^2 x = \frac{3}{4}

Take the square root of both sides: cosx=±32\cos x = \pm \frac{\sqrt{3}}{2}

Now, solve for xx. The solutions for cosx=32\cos x = \frac{\sqrt{3}}{2} are: x=2kπ±π6,kZx = 2k\pi \pm \frac{\pi}{6}, \quad k \in \mathbb{Z}

And for cosx=32\cos x = -\frac{\sqrt{3}}{2}: x=2kπ±5π6,kZx = 2k\pi \pm \frac{5\pi}{6}, \quad k \in \mathbb{Z}


Summary of Solutions:

  • a) x=kπ±π3,kZx = k\pi \pm \frac{\pi}{3}, \quad k \in \mathbb{Z}
  • b) x=2kπ±π6x = 2k\pi \pm \frac{\pi}{6} and x=2kπ±5π6,kZx = 2k\pi \pm \frac{5\pi}{6}, \quad k \in \mathbb{Z}

Would you like further clarification or more details?


Here are 5 related questions for further exploration:

  1. How do we derive the general solution for cosine equations?
  2. What is the significance of the trigonometric identities used here?
  3. Can cosine equations with different angles, like cos(2x)\cos(2x), have the same periodicity as cos(x)\cos(x)?
  4. How do sine and cosine equations differ in terms of their general solutions?
  5. What are some real-world applications of solving trigonometric equations?

Tip: When solving trigonometric equations, always check for all possible solutions within the given interval, as cosine and sine functions are periodic.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Cosine Function
Algebraic Equations

Formulas

cos(θ) = cos(2kπ ± θ)
General solution for cos(θ) = cos(α)
Quadratic equation in terms of cosine: 4cos^2(x) - 3 = 0

Theorems

Cosine identity
Trigonometric equation solving methods
Quadratic equation in trigonometry

Suitable Grade Level

Grades 10-12