Problem 37 - Algebra 2

Question

Solve each equation for all values of \(\theta\) if \(\theta\) is measured in radians. \(2 \sin ^{2} \theta+(\sqrt{2}-1) \sin \theta=\frac{\sqrt{2}}{2}\)

Step-by-Step Solution

Answer

Solve the quadratic and map the solutions to possible \(\theta\).

Step 1: Rewrite the Equation

First, notice the equation: \[ 2 \sin^2 \theta + (\sqrt{2} - 1) \sin \theta = \frac{\sqrt{2}}{2} \]We will rearrange all terms to form a quadratic equation in terms of \( \sin \theta \). Subtract \( \frac{\sqrt{2}}{2} \) from both sides to get:\[ 2 \sin^2 \theta + (\sqrt{2} - 1) \sin \theta - \frac{\sqrt{2}}{2} = 0 \]

Step 2: Substitute and Set as Quadratic

Let \( x = \sin \theta \). The equation becomes:\[ 2x^2 + (\sqrt{2} - 1)x - \frac{\sqrt{2}}{2} = 0 \]This is now a quadratic equation in the standard form \( ax^2 + bx + c = 0 \).

Step 3: Use the Quadratic Formula

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). For our equation: - \( a = 2 \)- \( b = \sqrt{2} - 1 \)- \( c = -\frac{\sqrt{2}}{2} \)Substitute these values into the formula:\[ x = \frac{-(\sqrt{2} - 1) \pm \sqrt{(\sqrt{2} - 1)^2 - 4(2)(-\frac{\sqrt{2}}{2})}}{4} \]

Step 4: Simplify the Discriminant

Calculate \( (\sqrt{2} - 1)^2 \): - \( (\sqrt{2} - 1)^2 = 2 - 2\sqrt{2} + 1 = 3 - 2\sqrt{2} \)Calculate the discriminant: \[ 3 - 2\sqrt{2} + 4 \cdot 2 \cdot \frac{\sqrt{2}}{2} = 3 - 2\sqrt{2} + 4\sqrt{2} = 3 + 2\sqrt{2} \]

Step 5: Continue Solving with Simplified Discriminant

Now, substitute the simplified discriminant back into the quadratic formula:\[ x = \frac{-(\sqrt{2} - 1) \pm \sqrt{3 + 2\sqrt{2}}}{4} \] Solving this gives possible values for \( x \), the \( \sin \theta \).

Step 6: Find Solutions for \(\theta\)

Recall that \( x = \sin \theta \). Both roots found from the quadratic formula provide potential \( \sin \theta \) values.Calculate \( \theta \) for the possible values of \( \sin \theta \) using all applicable solutions within the range \( \theta \in [0, 2\pi) \).

Key Concepts

Trigonometric Equations

Trigonometric equations involve trigonometric functions, like sine and cosine, that need to be solved for specific angles. These problems often require the use of algebraic manipulation. For instance, you need to rearrange terms and use identities like the Pythagorean identity. In the exercise, the trigonometric equation is expressed in terms of the sine function:- The original equation is \(2 \sin^2 \theta + (\sqrt{2} - 1) \sin \theta = \frac{\sqrt{2}}{2}\).- The task is to solve this for all angle values \(\theta\) measured in radians.The problem involves converting a trigonometric equation into a quadratic equation, making it simpler to solve using algebraic methods. Understanding these methods is crucial to finding solutions to trigonometric equations effectively.

Sin Function

The sine function relates the angle of a right triangle to the ratio of the length of the opposite side over the hypotenuse. It is a periodic function, meaning it repeats its values in regular intervals. In trigonometry, the sine function is vital for solving equations where the angle or arc is unknown.- In our equation, \(\sin \theta\) is squared and multiplied by constants.- It's crucial to isolate \(\sin \theta\) to simplify and solve the equation.In complex equations, \(\sin \theta\) might be treated as a variable (as done in the exercise with \(x = \sin \theta\)). By solving for \(x\), you find the potential values of the sine function, providing possible solutions for \(\theta\). Understanding the sine function’s domain and range is essential when considering valid solutions.

Quadratic Formula

The quadratic formula is a key tool for solving quadratic equations of the form \(ax^2 + bx + c = 0\). It provides solutions in the form:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula is used extensively in mathematics to find the roots of quadratic equations. In the given trigonometric equation, the sine function leads to a quadratic form articulated as:- \(2x^2 + (\sqrt{2} - 1)x - \frac{\sqrt{2}}{2} = 0\), where \(x\) is \(\sin \theta\).By inserting values for \(a\), \(b\), and \(c\) in the quadratic formula, one can compute possible values for \(x\). These roots are then translated back into the sine function, allowing us to find the corresponding angles \(\theta\). Understanding this formula aids in resolving quadratic expressions resulting from trigonometric equations.

Angle Measurement in Radians

Angles can be measured in degrees or radians, with radians being the standard unit in mathematics for calculus and trigonometry. One full circle is \(2\pi\) radians, which equals 360 degrees. Radians provide a direct link between the study of circles and angles.- In this exercise, angles \(\theta\) are measured in radians.Radians are especially useful for working with the periodic nature of trigonometric functions since they simplify the derivatives and integrals of these functions. When solving equations like \(2 \sin^2 \theta + (\sqrt{2} - 1) \sin \theta = \frac{\sqrt{2}}{2}\), understanding radians helps in determining the solutions within the interval \([0, 2\pi)\). This interval encapsulates all possible solutions for a single periodic cycle of sine and cosine functions.