Solve Trigonometric Equations Requiring The Quadratic Formula (Give All Answers, In (0,2 ), In Radians
Solving trigonometric equations is a fundamental skill in mathematics, particularly in calculus, physics, and engineering. Some equations involve quadratic expressions of trigonometric functions, requiring the use of the quadratic formula to find solutions. When tackling these equations, it is essential to understand how to manipulate the equations into quadratic form, apply the quadratic formula correctly, and interpret the solutions within the specified interval—in this case, between 0 and 2 radians. This comprehensive guide will walk you through the process of solving such equations, providing step-by-step solutions, key concepts, and tips to master this topic.
Understanding the Quadratic Form in Trigonometric Equations
Many trigonometric equations can be rewritten into a quadratic form by substitution. For example, equations involving sin θ or cos θ can often be expressed as quadratic equations in terms of these functions.
Common Patterns Leading to Quadratic Equations
- Equations like sin²θ + a sin θ + b = 0
- Equations like cos²θ + c cos θ + d = 0
- Mixed forms that can be converted into quadratic form through substitution
Why Use the Quadratic Formula?
The quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
provides a direct method to solve quadratic equations when factoring is difficult or impossible. When the quadratic form appears in trigonometric equations, applying this formula helps find all possible solutions for the trigonometric functions, which we then convert back to the original variable, θ.
Step-by-Step Process for Solving Trigonometric Equations Using the Quadratic Formula
- Rewrite the equation in terms of a single trigonometric function
- Express the equation in quadratic form
- Identify coefficients a, b, and c
- Calculate the discriminant \( D = b^2 - 4ac \)
- Determine the nature of solutions based on the discriminant
- Apply the quadratic formula to find the solutions for the trigonometric function
- Solve for θ within the interval (0, 2) radians
- Verify solutions and discard extraneous roots outside the interval
Example 1: Solving a Basic Trigonometric Quadratic Equation
Suppose we want to solve:
\[ \sin^2 θ - \sin θ - 2 = 0 \quad \text{for} \quad θ \in (0, 2) \]
Step 1: Substitution
Let \( x = \sin θ \). The equation becomes:
\[ x^2 - x - 2 = 0 \]
Step 2: Quadratic form
Already in quadratic form with \( a=1 \), \( b=-1 \), \( c=-2 \).
Step 3: Calculate discriminant
\[ D = (-1)^2 - 4 \times 1 \times (-2) = 1 + 8 = 9 \]
Since \( D>0 \), there are two real solutions.
Step 4: Apply quadratic formula
\[
x = \frac{-(-1) \pm \sqrt{9}}{2 \times 1} = \frac{1 \pm 3}{2}
\]
Solutions:
- \( x = \frac{1 + 3}{2} = 2 \)
- \( x = \frac{1 - 3}{2} = -1 \)
Step 5: Interpret solutions for \( \sin θ \)
Recall that \( \sin θ \) must be in [-1, 1].
- \( \sin θ = 2 \) → Not possible, since sine cannot be greater than 1.
- \( \sin θ = -1 \) → Possible.
Step 6: Find θ in (0, 2) for \( \sin θ = -1 \)
\[
\sin θ = -1 \quad \Rightarrow \quad θ = \frac{3π}{2} \approx 4.712 \text{ radians}
\]
But we are asked for solutions in (0, 2) radians. Since \( \frac{3π}{2} \approx 4.712 \), which exceeds 2 radians, no solutions within the interval.
Result: No solutions in (0, 2) radians for the given equation.
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Example 2: Solving a Cosine-Based Equation
Let's consider:
\[ 2 \cos^2 θ - 3 \cos θ + 1 = 0 \quad \text{for} \quad θ \in (0, 2) \]
Step 1: Substitution
Let \( x = \cos θ \). The quadratic becomes:
\[ 2x^2 - 3x + 1 = 0 \]
Step 2: Identify coefficients
\( a=2 \), \( b=-3 \), \( c=1 \).
Step 3: Calculate discriminant
\[ D = (-3)^2 - 4 \times 2 \times 1 = 9 - 8 = 1 \]
Step 4: Find roots
\[
x = \frac{3 \pm \sqrt{1}}{2 \times 2} = \frac{3 \pm 1}{4}
\]
Solutions:
- \( x = \frac{3 + 1}{4} = 1 \)
- \( x = \frac{3 - 1}{4} = \frac{1}{2} \)
Step 5: Solutions for \( \cos θ \)
- \( \cos θ = 1 \)
- \( \cos θ = \frac{1}{2} \)
Step 6: Find θ in (0, 2)
- For \( \cos θ = 1 \):
\[
\cos θ = 1 \quad \Rightarrow \quad θ = 0
\]
But 0 is not in the interval (0, 2), so discard.
- For \( \cos θ = \frac{1}{2} \):
\[
θ = \frac{π}{3} \approx 1.047 \text{ radians}
\]
and
\[
θ = 2π - \frac{π}{3} = \frac{5π}{3} \approx 5.236 \text{ radians}
\]
Since the interval is (0, 2), only \( θ = \frac{π}{3} \) ≈ 1.047 radians fits.
Result: The only solution in (0, 2) radians is approximately 1.047 radians.
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Special Considerations When Using the Quadratic Formula
- Discriminant Analysis:
- \( D > 0 \) indicates two real solutions.
- \( D = 0 \) indicates one real solution.
- \( D < 0 \) indicates no real solutions for the trigonometric function, hence no solutions in the interval.
- Range of Trigonometric Functions:
- Interval of Solutions:
General Strategy for Complex Equations
- Convert the equation into quadratic form using substitution.
- Calculate the discriminant and analyze the roots.
- For each valid root within [-1, 1], find the corresponding θ using inverse trig functions.
- Use the periodicity of sine and cosine to find all solutions within (0, 2).
- Discard extraneous solutions outside the interval.
Additional Examples and Practice Problems
To solidify understanding, here are practice problems:
- Solve \( 3 \sin^2 θ - 2 \sin θ - 1 = 0 \) for \( θ \in (0, 2) \).
- Solve \( \cos^2 θ + \cos θ - 2 = 0 \) for \( θ \in (0, 2) \).
- Determine all solutions for \( 4 \sin^2 θ - 3 = 0 \) in \( (0, 2) \).
Working through these problems using the methods detailed above will enhance your ability to solve quadratic trigonometric equations efficiently and accurately.
Conclusion
Solving trigonometric equations involving quadratic expressions requires a systematic approach—rewriting the equations in quadratic form, applying the quadratic formula, and carefully analyzing the solutions within the specific interval. Remember to verify solutions against the range of the trigonometric functions and to consider the periodic nature of
