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Chapter 6: Problem 27

For each of the following equations, solve for (a) all radian solutions and (b) \(t\) if \(0 \leq t<2 \pi\). Give all answers as exact values in radians. Do not use a calculator. $$3 \sin t+5=-2 \sin t$$

Short Answer

Expert verified
For all solutions: \(t = \frac{3\pi}{2} + 2k\pi\); for \([0, 2\pi)\): \(t = \frac{3\pi}{2}\).

Step by step solution

01

Rearrange the Equation

Given the equation: \(3 \sin t + 5 = -2 \sin t\). To find the solutions, first, get all the \(\sin t\) terms on one side. Add \(2 \sin t\) to both sides: \(3 \sin t + 2 \sin t + 5 = 0\). This simplifies to \(5 \sin t + 5 = 0\).
02

Solve for \(\sin t\)

Subtract 5 from both sides to isolate the \(\sin t\) term: \(5 \sin t = -5\). Then, divide both sides by 5 to solve for \(\sin t\): \(\sin t = -1\).
03

Determine Radian Solutions for \(\sin t = -1\)

The sine function equals \(-1\) at specific points on the unit circle. For \(\sin t = -1\), this happens when \(t = \frac{3\pi}{2} + 2k\pi\), where \(k\) is any integer.
04

Find Solutions in the Interval \([0, 2\pi)\)

In the interval \([0, 2\pi)\), we only consider values that fit within these bounds. We know that \(\sin t = -1\) for \(t = \frac{3\pi}{2}\). Since \(\frac{3\pi}{2}\) is within \([0, 2\pi)\), it is the only solution within the specified range.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Sine Function
The sine function, often written as \(\sin t\), is a fundamental concept in trigonometry. It specifically deals with the ratio of the length of the opposite side to the hypotenuse in a right triangle. For the unit circle, which we'll talk about soon, \(\sin t\) represents the y-coordinate of a point on the circle.

Key aspects to remember about the sine function are:
  • The sine function oscillates between -1 and 1.
  • It is periodic with a period of \(2\pi\), meaning it repeats its values every \(2\pi\) units.
  • Important points to know by heart are where \(\sin t = 0\), \(\sin t = 1\), and \(\sin t = -1\).

For example, \(\sin t = 1\) at \(t = \frac{\pi}{2} + 2k\pi\), and \(\sin t = -1\) at \(t = \frac{3\pi}{2} + 2k\pi\). These are crucial for solving trigonometric equations swiftly.
The Unit Circle and its Importance
The unit circle is a circle with a radius of 1 centered at the origin of a coordinate plane. It's a handy tool for understanding trigonometric functions, including the sine function.

Here's why the unit circle is so crucial:
  • The coordinates of any point on the unit circle are \((\cos t, \sin t)\), linking both sine and cosine functions to angles.
  • The unit circle explains why trigonometric functions are periodic, since moving around the circle will eventually return you to the same point after traversing \(2\pi\) radians.
  • On the unit circle, \(\sin t\) equals the y-coordinate, which directly reflects how the sine function behaves as it moves up and down through values from -1 to 1.

By understanding the unit circle, you can easily determine when sine and cosine functions equal specific values, such as finding where \(\sin t = -1\), using the circle's properties to identify angles.
Radian Solutions in Trigonometric Equations
Radian solutions are the angles measured in radians that satisfy a trigonometric equation. In trigonometry, radian measures offer a more natural way of expressing angles compared to degrees.

Here's why radians are preferred:
  • Radians provide a direct way to relate the angle to the radius of the circle, making calculations involving arc lengths and areas more straightforward.
  • A complete circle in radian measures equates to \(2\pi\), which simplifies the period of trigonometric functions to more intuitive expressions.
  • Identifying solutions like \(\sin t = -1\) is often simpler in radians, as these measures directly link to commonly known points on the unit circle, such as \(t = \frac{3\pi}{2}\).

For trigonometric equations, solutions in radians become essential. These allow you to easily find precise answers, both in overall solutions (like \(t = \frac{3\pi}{2} + 2k\pi\)) and within specified ranges (like choosing only \(t = \frac{3\pi}{2}\) when limited to \([0, 2\pi)\)).

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Most popular questions from this chapter

For each equation, find all degree solutions in the interval \(0^{\circ} \leq \theta<360^{\circ}\). If rounding is necessary, round to the nearest tenth of a degree. Use your graphing calculator to verify each solution graphically. $$ 18 \sec ^{2} \theta-17 \tan \theta \sec \theta-12=0 $$

$$ \text { Find all solutions in radians. Approximate your answers to the nearest hundredth. } $$ $$ 15-9 \cos (6 t-5)=11 $$

In solving \(\csc \theta-2 \cot \theta=0\), one of the steps involves solving which equation? a. \(\sin \theta=1\) b. \(\sin \theta=-1\) c. \(\cos \theta=-\frac{1}{2}\) d. \(\cos \theta=\frac{1}{2}\)

Solve each equation for \(\theta\) if \(0^{\circ} \leq \theta<360^{\circ}\). $$ 2 \sin \theta+\sin 2 \theta=0 $$

The problems that follow review material we covered in Section \(5.4\). If \(\sin A=\frac{2}{3}\) with \(A\) in the interval \(0^{\circ} \leq A \leq 90^{\circ}\), find $$ \cos \frac{A}{2} $$
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