Problem 31 - Algebra 2

Question

Solve each equation in the interval from 0 to 2\(\pi .\) Round your answers to the nearest hundredth. \(-3 \sin 2 \theta=1.5\)

Step-by-Step Solution

Answer

After computing the solutions, we find the angles \( \theta\) that meet the requirements of the original problem are: \( \theta \approx 3.93\) and \( \theta \approx 5.64\) in the range of 0 to \(2 \pi\).

Step 1: Simplify the Equation

First, you need to isolate \(\sin 2 \theta\) on one side of the equation. We can do this by dividing each side of the equation by -3. Doing this, we find: \[\sin 2 \theta = -0.5\]

Step 2: Solve for \(2 \theta\)

Now, we apply the inverse sine function to both sides of the equation. This gives us two possible solutions for \(2\theta\), because sine function repeats after an interval of \(\pi\). The equation is then: \[2 \theta = \arcsin(-0.5), 2\theta = \pi - \arcsin(-0.5)\]

Step 3: Solve for \( \theta\)

We can now solve for \( \theta\) by dividing each side of the equation by 2. The two equations become: \[\theta = \frac{\arcsin(-0.5)}{2}, \theta = \frac{\pi - \arcsin(-0.5) }{2}\] Note that each of the above equations gives us more than one value of \(\theta\), because the sine of a angle and its supplement is same, the values range from 0 to \(2 \pi\).

Step 4: Find all Possible Values of \(\theta\)

Now, calculate the values from the last step and search for values of \(\theta\) in the range 0 to \(2 \pi\). Linke mentioned before, the sine function is cyclic, so you should find two possible angles within the desired range that solve the original equation

Key Concepts

Sine Function

The sine function is an essential part of trigonometry and is widely used in various fields such as physics, engineering, and mathematics. Sine, denoted as \( \sin \), is a non-linear function that takes an angle as an input and gives a ratio of two sides of a right triangle. Specifically, it is the ratio of the length of the side opposite to the angle over the hypotenuse.

• Definition: Given a right triangle, \( \sin(\theta) = \frac{\text{opposite side}}{\text{hypotenuse}} \)
• Range: The values of sine function range from -1 to 1, meaning \(-1 \leq \sin(\theta) \leq 1\).
• Periodicity: Sine is periodic with a period of \(2\pi\), which means it repeats every \(2\pi\) radians.

The sine function is part of the circle of trigonometric functions and can be visualized using the unit circle where the angle's sine is the \(y\)-coordinate of the point on this circle.

Inverse Trigonometric Functions

Inverse trigonometric functions allow us to find the angle that corresponds to a given trigonometric value. The inverse sine, known as arcsine and denoted as \( \arcsin \), returns the angle \( \theta \) for a given sine value.

• Principal Value: The principal value of arcsine returns angles only in the interval \([-\frac{\pi}{2}, \frac{\pi}{2}]\).
• Multiple Solutions: Since sine is periodic with multiple angles having the same sine value, we create additional solutions by using the fact that \( \sin(\pi - \theta) = \sin(\theta) \).

For example, if \( \sin(2\theta) = -0.5 \), apply the inverse function to get \(2\theta = \arcsin(-0.5)\). In agreement with periodicity, we consider the supplemental angle \( \pi - \arcsin(-0.5) \).

Angle Measurement

In mathematics, angles are typically measured in radians or degrees. When dealing with trigonometric equations, radians are preferred because they offer a natural way to describe angle measures based on pi.

• Radians: One radian is the angle created when the length of the arc of a circle is equal to the radius of the circle. There are \(2\pi\) radians in a full circle.
• Conversion: We can convert between degrees and radians using the relations \(180^\circ = \pi\) radians.

This understanding is critical for interpreting the solutions of trigonometric equations and ensuring the solutions lie within the specified interval \([0, 2\pi]\).

Mathematical Intervals

Mathematical intervals define a range of values that a variable can take. They are particularly useful in defining the solution set for trigonometric equations.

• Closed Interval \([a, b]:\) Includes both endpoints; all values \(x\) such that \(a \leq x \leq b\).
• Open Interval \((a, b):\) Does not include the endpoints; all values \(x\) such that \(a < x < b\).

In the original problem, we sought solutions in the interval from \(0\) to \(2\pi\). This closed interval represents all possible angles within one full rotation around a circle in radian measure. Identifying solutions within these bounds ensures all outcomes are valid under given conditions.