Problem 58 - Trigonometry

Question

Find all degree solutions. $$ 3 \sin ^{2} 2 \theta-2 \sin 2 \theta-5=0 $$

Step-by-Step Solution

Answer

The solutions are \(\theta = 135^\circ + k \cdot 180^\circ\) with integer \(k\).

Step 1: Simplify the Equation

We are given the equation: \[3 \sin^2 2\theta - 2 \sin 2\theta - 5 = 0.\]Let's set \(x = \sin 2\theta\). So, the equation becomes:\[3x^2 - 2x - 5 = 0.\] This is a quadratic equation in terms of \(x\).

Step 2: Solve the Quadratic Equation

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = -2\), and \(c = -5\), calculate:\[x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-5)}}{2 \cdot 3}.\]Simplify inside the square root:\[x = \frac{2 \pm \sqrt{4 + 60}}{6} = \frac{2 \pm \sqrt{64}}{6} = \frac{2 \pm 8}{6}.\]This gives us the solutions:\[x_1 = \frac{2 + 8}{6} = \frac{10}{6} = \frac{5}{3},\]\[x_2 = \frac{2 - 8}{6} = \frac{-6}{6} = -1.\]

Step 3: Solve for \(\theta\) using \(x_2\)

For \(x_2 = -1\), we have \(\sin 2\theta = -1\). The solution for this, considering the sine function, is:\[2\theta = 270^\circ + k \cdot 360^\circ,\]where \(k\) is an integer. Therefore, solving for \(\theta\):\[\theta = 135^\circ + k \cdot 180^\circ.\]

Step 4: Evaluate \(x_1\) for Real Solutions

For \(x_1 = \frac{5}{3}\), we have \(\sin 2\theta = \frac{5}{3}\). However, the sine function must be within the range \([-1, 1]\). Since \(\frac{5}{3} > 1\), there are no real solutions for this case.

Step 5: List All Degree Solutions

The valid solutions from step 3 are:\[\theta = 135^\circ + k \cdot 180^\circ,\]where \(k\) is an integer. These solutions cover all the possible angles that satisfy the original equation, keeping in mind the periodic nature of the sine function.

Key Concepts

Quadratic Equations

Quadratic Equations are commonly found in many areas of mathematics and science. They typically appear in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( x \) is the variable. A crucial part of solving these equations is understanding the quadratic formula:

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

This formula allows us to calculate the possible values of \( x \) that satisfy the equation. For successful application, one should:

• Identify the coefficients \( a \), \( b \), and \( c \) from the equation.
• Substitute these values into the formula accurately.

When the values are plugged in, calculations within the square root (discriminant) signify the nature of the roots:

• If the discriminant \( b^2 - 4ac \) is positive, there are two real solutions.
• If it is zero, there is one real solution.
• If it's negative, there are no real solutions, leading to complex numbers.

In our equation example, substitutions and simplifications lead to finding both relevant and irrelevant solutions based on the sine function's range constraints.

Sine Function

The Sine Function is an essential trigonometric function, often denoted as \( \sin \), that relates an angle in a right-angled triangle to a ratio of sides. When concerning angles, it is defined as the ratio of the length of the opposite side to the hypotenuse. For values outside a triangle context, it considers the coordinates of points on a unit circle:

• \( \sin \theta = \text{opposite} / \text{hypotenuse} \)

The sine function oscillates between -1 and 1, repeating every 360 degrees (or \(2\pi\) radians). This characteristic is crucial when solving equations like the one provided. Typically:

• \( \sin x = 0 \) at every 180 degree interval \( (0, 180, 360,...) \)
• Maximum value \( \sin x = 1 \) occurs at 90 degrees.
• Minimum value \( \sin x = -1 \) occurs at 270 degrees.

Understanding this periodical behavior helps in deciphering the possible solutions, ensuring they remain within the naturally occurring range of the function: from -1 to 1. Moreover, it highlights why solutions like \( \sin 2\theta = \frac{5}{3} \) are invalid, as they fall outside this inherent range.

Degree Solutions

When solving trigonometric equations, finding Degree Solutions involves determining the angles that satisfy the equation within the specific bounds of a full circle, or 360 degrees. With angles in trigonometry:

• We consider the periodicity, since the sine, cosine, and tangent functions repeat their values in regular intervals.
• Multiple "equivalent" solutions are possible due to this periodic nature.
• For our equation, the focus is on solutions of the form \( \theta = 135^\circ + k \cdot 180^\circ \), where \( k \) is any integer that captures all solutions over time.

This formula essentially adds full revolutions \( (360^\circ) \) divided by 2 (since \(2\theta\) was solved, then halved) to an anchor point \( (135^\circ) \). Each full rotation represents a complete cycle of the sine function's waveform, ensuring comprehensive coverage of all potential solutions within the radians specified. Understanding degree solutions requires the flexibility of imagining the rotation and repetitive nature of angles within circles.