Math Problem Statement

Berechne alle Lösungen a

Berechne alle Lösungen a

Solution

To solve the equation provided:

(21+tan2(α))2+151+tan2(α)=4,\left(\frac{2}{1 + \tan^2(\alpha)}\right)^2 + \frac{15}{1 + \tan^2(\alpha)} = 4,

we can proceed by simplifying the expression and solving for α\alpha.

Step-by-Step Solution:

  1. Introduce a Substitution: Let x=1+tan2(α)x = 1 + \tan^2(\alpha). This substitution uses the Pythagorean identity 1+tan2(α)=sec2(α)1 + \tan^2(\alpha) = \sec^2(\alpha).

    With this substitution, the equation becomes: (2x)2+15x=4.\left(\frac{2}{x}\right)^2 + \frac{15}{x} = 4.

  2. Simplify the Equation: Expanding (2x)2\left(\frac{2}{x}\right)^2 gives 4x2\frac{4}{x^2}, so the equation becomes: 4x2+15x=4.\frac{4}{x^2} + \frac{15}{x} = 4.

  3. Clear the Denominator by Multiplying by x2x^2: Multiply both sides by x2x^2 to eliminate the fractions: 4+15x=4x2.4 + 15x = 4x^2.

  4. Rearrange the Equation into a Standard Quadratic Form: Bring all terms to one side of the equation: 4x215x4=0.4x^2 - 15x - 4 = 0.

  5. Solve the Quadratic Equation: Use the quadratic formula: x=(15)±(15)244(4)24.x = \frac{-(-15) \pm \sqrt{(-15)^2 - 4 \cdot 4 \cdot (-4)}}{2 \cdot 4}. Simplify the expression: x=15±225+648.x = \frac{15 \pm \sqrt{225 + 64}}{8}. [

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Algebra
Quadratic Equations

Formulas

1 + tan^2(α) = sec^2(α)
Quadratic equation formula: ax^2 + bx + c = 0

Theorems

Pythagorean identity
Quadratic formula

Suitable Grade Level

Grades 10-12