Question

Consider the trigonometric equation \(\sin ^{2} \theta+\sin \theta-1=0\) a) Can you solve the equation by factoring? b) Use the quadratic formula to solve for \(\sin \theta\) c) Determine all solutions for \(\theta\) in the interval \(0 < \theta \leq 2 \pi .\) Give answers to the nearest hundredth of a radian, if necessary.

Short answer

\(\theta = 0.61\) or \(\theta = 2.53\)

Step-by-step solution

- Identify the type of equation

Notice that the equation \(\sin^2 \theta + \sin \theta - 1 = 0\) resembles a quadratic equation of the form \(ax^2 + bx + c = 0\) with \(x = \sin \theta\).

- Attempt factoring

Rewrite the equation as \(\sin^2 \theta + \sin \theta - 1 = 0\). Try to factor it: \([\sin \theta + 1][\sin \theta - 1] = 0\). However, this does not satisfy the original equation. Therefore, factoring is not successful.

- Apply the quadratic formula

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1\), \(b = 1\), and \(c = -1\). Plug these values into the formula to get: \(\sin \theta = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2}\).

- Solve for \(\sin \theta\)

Calculate the solutions: \(\sin \theta = \frac{-1 + \sqrt{5}}{2}\) and \(\sin \theta = \frac{-1 - \sqrt{5}}{2}\). Given that the range of \(\sin \theta\) is \([-1, 1]\), only \(\sin \theta = \frac{-1 + \sqrt{5}}{2} \) is valid.

- Find \(\theta \)

Using the valid \(\sin \theta\) value, solve for \(\theta\) in the interval \(0 < \theta \leq 2 \pi\). Use the arcsin function: \(\theta = \arcsin \left( \frac{-1 + \sqrt{5}}{2} \right)\) and \(\theta = \pi - \arcsin \left( \frac{-1 + \sqrt{5}}{2} \right)\). Calculate the numerical values for \(\theta\).

- Numerical approximation

Approximate the solutions to the nearest hundredth. Using a calculator: \(\theta \approx 0.61 \) and \(\theta \approx 2.53\).

Key concepts

quadratic formula

The quadratic formula is a powerful tool for solving quadratic equations. A quadratic equation is generally in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. The quadratic formula helps to find the values of \(x\) that satisfy the equation. The formula is represented as: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). In our equation, \(\text{sin}^2 \theta + \text{sin} \theta - 1 = 0\), we can identify \(a = 1\), \(b = 1\), and \(c = -1\). Plugging these into the quadratic formula, we get two potential solutions for \(\text{sin} \theta\): \(\text{sin} \theta = \frac{-1 + \sqrt{5}}{2}\) and \(\text{sin} \theta = \frac{-1 - \sqrt{5}}{2}\). Since \(\text{sin} \theta\) must be within the interval \([-1, 1]\), only the first value is valid.

trigonometric functions

Trigonometric functions are fundamental in understanding angles and their relationships in geometry. The key functions include \(\text{sin} ( \theta )\), \(\text{cos} ( \theta )\), and \(\text{tan} ( \theta )\). In this context, we solved an equation involving \(\text{sin} (\theta)\). Importantly, the sine function gives the ratio of the opposite side to the hypotenuse in a right-angle triangle. Knowing the value of \(\text{sin} (\theta)\) enables us to find \(\theta\) within given intervals using \(\text{arcsin}\), which is the inverse sine function that helps to determine the angle when its sine value is known.

interval notation

Interval notation is a way of writing subsets of the real number line. It is especially useful in describing the set of values of \(\theta\) we need. For our purposes, we only consider \(\theta\) within the range \(0 < \theta \text{leq} 2 \text{pi}\). This interval describes angles between \(0\) and \(2 \text{pi}\) but not including \(0\), signifying a complete cycle in the unit circle. This ensures we capture all possible angles within one full rotation around the circle where the solution may lie.

numerical solutions

Numerical solutions involve approximating values when exact figures are complex. In our solution, we calculated the angles \(\theta\) using numerical methods or a calculator. \(\text{sin} \theta = \frac{-1 + \sqrt{5}}{2}\) approximates to \(0.61\) and \(\text{pi} - \arcsin \left( \frac{-1 + \sqrt{5}}{2} \right)\) approximates to \(2.53\). Such approximations are practical, especially when exact answers are irrational numbers, ensuring that answers are easy to read and understand while maintaining sufficient accuracy for practical use.