Problem 62 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 39-62: Find the solutions of the equation that are in the interval \([0,2 \pi\) ). $$ 2 \sin v \csc v-\csc v=4 \sin v-2 $$

Step-by-Step Solution

Answer

The given equation has no real solutions in the interval \([0, 2\pi)\).

Step 1: Simplify the LHS of the equation

The given equation is \(2 \sin v \csc v - \csc v = 4 \sin v - 2\). Start by simplifying the left-hand side (LHS). Notice that \(\csc v = \frac{1}{\sin v}\). Thus, \[2 \sin v \csc v - \csc v = 2 - \frac{1}{\sin v}.\]

Step 2: Rearrange the equation

Rearrange the equation so all terms are on one side:\[2 - \frac{1}{\sin v} = 4 \sin v - 2.\] Add \(\frac{1}{\sin v}\) to both sides to form: \[2 = 4 \sin v - 2 + \frac{1}{\sin v}.\]

Step 3: Combine terms on the RHS

Combine like terms on the right-hand side (RHS) of the equation:\[4 \sin v - 2 + \frac{1}{\sin v} = \frac{4 \sin^2 v - 2 \sin v + 1}{\sin v}.\] The equation is now:\[2 \sin v = \frac{4 \sin^2 v - 2 \sin v + 1}{\sin v}.\]

Step 4: Solve for \(\sin v\)

Let's simplify and solve for \(\sin v\). Multiply through by \(\sin v\) to clear the fraction:\[2 \sin^2 v = 4 \sin^2 v - 2 \sin v + 1.\] Rearrange the terms to form the quadratic equation: \[2 \sin^2 v - 2 \sin v + 1 = 0.\]

Step 5: Solve the quadratic equation

To solve the quadratic equation \(2 \sin^2 v - 2 \sin v + 1 = 0\), apply the quadratic formula:\[\sin v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},\] where \(a = 2\), \(b = -2\), and \(c = 1\). Calculate:\[\sin v = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(2)(1)}}{2(2)} = \frac{2 \pm \sqrt{4 - 8}}{4}.\] This simplifies to \(\sin v = \frac{2 \pm \sqrt{-4}}{4}\). Since \(\sqrt{-4}\) does not yield real solutions, conclude there are no real solutions.

Key Concepts

Quadratic Equations

Quadratic equations are a type of polynomial equation of degree 2, often written in the form:

• \( ax^2 + bx + c = 0 \)

where \( a \), \( b \), and \( c \) are constants, and \( x \) represents the unknown variable.
Solving quadratic equations is a fundamental skill in algebra and involves finding values of \( x \) that make the equation true.One common method to solve quadratic equations is using the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

This formula requires calculating the discriminant, \( b^2 - 4ac \).
The nature of the roots, whether real or complex, depends on the discriminant:

• If the discriminant is positive, there are two distinct real solutions.
• If it is zero, there is exactly one real solution.
• If it is negative, the solutions are complex (not real).

In the exercise provided, after simplifying the trigonometric equation, a quadratic equation in terms of \( \sin v \) was formed.
However, the discriminant was negative, leading to no real solutions.

Interval Solutions

Interval solutions involve finding specific values within a given range that satisfy an equation.
Here, the range, or interval, is \([0, 2\pi)\), which represents all possible angles from 0 to just under \(2\pi\) radians (or 0 to 360 degrees).
This is a common consideration when solving trigonometric equations.To find solutions in a specific interval:

• The solutions must be calculated or checked to fall within the given boundaries.
• The endpoints are usually open or closed to indicate whether they are included in the set of solutions.

For the exercise at hand, any possible solutions for \(v\) would need to fall within \([0, 2\pi)\).
However, because the quadratic equation produced no real solutions, there are no corresponding angles \(v\) within the interval that satisfy the equation.

Trigonometric Identities

Trigonometric identities are equations that relate the trigonometric functions like sine, cosine, and tangent.
They are true for all values of the included variables for which both sides of the equation are defined.
These identities are used extensively to simplify trigonometric expressions and solve trigonometric equations.In the provided exercise, some key trigonometric identities played a role:

• The identity \( \csc v = \frac{1}{\sin v} \) was used to help simplify both sides of the equation.
• This simplification helped to transform the trigonometric equation into a more manageable quadratic form in terms of \( \sin v \).

Understanding and using trigonometric identities like these is crucial for simplifying equations and making them solvable.
They allow the transformation of complex expressions into simpler ones, often leading to the solution or making the equation easier to handle.