Problem 53 - A Graphical Approach to Precalculus with Limits

Question

Give solutions over the interval \([0,2 \pi)\) as approximations to the nearest hundredth when exact values cannot be determined. You may need to use the quadratic formula. Give approximate answers in Exercises \(59-64\) to the nearest tenth of a degree over the interval \(\left[0^{\circ}, 360^{\circ}\right)\) $$3 \sin ^{2} x-\sin x=2$$

Step-by-Step Solution

Answer

The solutions are \(90°\), \(251.6°\), and \(316.8°\).

Step 1: Set up the equation

Identify that the given equation is already in a quadratic form in terms of sin(x): \(3 \sin^2 x - \sin x - 2 = 0\). Let \(u = \sin x\). This simplifies the equation to \(3u^2 - u - 2 = 0\).

Step 2: Apply the quadratic formula

The quadratic formula is \(u = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\). Here \(a = 3\), \(b = -1\), and \(c = -2\). Substitute these values into the formula to find \(u\).

Step 3: Calculate the discriminant

Compute the discriminant \(b^2-4ac = (-1)^2 - 4 \cdot 3 \cdot (-2) = 1 + 24 = 25\). Since the discriminant is positive, there are two real solutions for \(u\).

Step 4: Solve for sin(x) values

Substitute into the quadratic formula: \(u = \frac{-(-1)\pm\sqrt{25}}{2\cdot3}\). Simplifying gives \(u = \frac{1\pm5}{6}\). The solutions are \(u_1 = 1\) and \(u_2 = -\frac{2}{3}\).

Step 5: Find x for each u value

For \(\sin x = 1\), the solution over \([0, 2\pi)\) is \(x = \frac{\pi}{2}\). For \(\sin x = -\frac{2}{3}\), use a calculator to find the arcsin. The solutions are approximately \(x \approx 4.39\) or \(x \approx 5.53\).

Step 6: Converting radians to degrees

Convert each angle from radians to degrees. \(\frac{\pi}{2}\) radians is 90°, approximately 4.39 radians is 251.6°, and approximately 5.53 radians is 316.8°.

Key Concepts

Solutions of Trigonometric Equations

When solving trigonometric equations like \(3\sin^2x - \sin x - 2 = 0\), it is crucial to find the values of \(x\) that satisfy the equation within a specific interval. In this exercise, we consider the interval \([0, 2\pi)\).
This involves several steps:

• First, the equation is transformed into a quadratic form, making it easier to solve. This is done by letting \(u = \sin x\), transforming it into a standard quadratic equation \(3u^2 - u - 2 = 0\).
• The next step is to solve this quadratic equation for \(u\). The solutions for \(u\) can then be used to find the corresponding values of \(x\).
• Trigonometric functions like sine have specific ranges and properties, which must be considered. For instance, \(\sin x\) ranges from \(-1\) to \(1\). Thus, valid solutions for \(u\) must also fall within this range.

Once these steps are completed, the problem transforms from solving a complex trigonometric equation to handling quadratic solutions. Understanding this transformation is key to mastering such equations.

Radians to Degrees Conversion

Converting between radians and degrees is an essential part of solving trigonometric problems in different units. Radians and degrees are two units used to measure angles, and each has its use in mathematics and science. Understanding this conversion is critical when interpreting or verifying solutions.
To convert from radians to degrees, use the conversion factor:

• 1 radian = \(\frac{180}{\pi}\) degrees.
• Thus, to convert an angle \(x\) in radians to degrees, calculate \(x \times \frac{180}{\pi}\).

For instance:

• The angle \(\frac{\pi}{2}\) in radians is equivalent to 90° because \(\frac{\pi}{2} \times \frac{180}{\pi} = 90\).
• An approximate angle like 4.39 radians converts to degrees as \(4.39 \times \frac{180}{\pi} \approx 251.6\) degrees.
• Similarly, 5.53 radians converts to \(5.53 \times \frac{180}{\pi} \approx 316.8\) degrees.

This conversion is not only important for finding solutions but also for interpreting them in practical scenarios, where degrees are often used.

Quadratic Formula

The quadratic formula is a powerful tool for solving equations of the form \(ax^2 + bx + c = 0\). For this exercise, it was applied to find the roots of \(3\sin^2x - \sin x - 2 = 0\). Knowing how to employ the quadratic formula effectively is crucial.
The formula itself is:\[\ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\]It provides the solutions to any quadratic equation, given the coefficients \(a\), \(b\), and \(c\):

• First, identify the values of \(a\), \(b\), and \(c\) from the equation. In this case, \(a=3\), \(b=-1\), and \(c=-2\).
• Compute the discriminant \(b^2 - 4ac\). A positive discriminant indicates two real and distinct solutions. For this equation, the discriminant is 25, which is positive.
• Use the quadratic formula to find the roots \(u_1 = 1\) and \(u_2 = -\frac{2}{3}\).

This process transforms a complex trigonometric problem into a straightforward algebraic solution of a quadratic equation.

Trigonometric Functions

Trigonometric functions like sine, cosine, and tangent are fundamental in describing relationships in triangles and circular functions. In this problem, the sine function is the main focus, as we solve \(3\sin^2x - \sin x - 2 = 0\). Understanding these functions is crucial:

• The sine function \(\sin x\) has a range of \([-1, 1]\). This range limits the possible solutions to the equation, as \(\sin x\) cannot exceed these bounds.
• The unit circle concept helps visualize trigonometric identities and values. For example, \(\sin x = 1\) corresponds to \(x = \frac{\pi}{2}\).
• Arc functions like \(\arcsin\) allow finding angles for given sine values. This is relevant when \(\sin x\) does not yield neat intersections, as with \(\sin x = -\frac{2}{3}\).

These principles together help solve and understand the behavior of trigonometric equations, enabling students to connect algebraic solutions with geometric interpretations.