/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 30 Solve the given trigonometric eq... [FREE SOLUTION] | 91Ó°ÊÓ

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Solve the given trigonometric equation exactly on \(0 \leq \theta<2 \pi\). $$\sin ^{2} \theta+2 \sin \theta+1=0$$

Short Answer

Expert verified
The solution is \(\theta = \frac{3\pi}{2}\).

Step by step solution

01

Recognize the quadratic form

The given equation \( \sin^2 \theta + 2 \sin \theta + 1 = 0 \) is a quadratic equation in terms of \( \sin \theta \). We recognize it by comparing it with the general form \( ax^2 + bx + c = 0 \), where \( x = \sin \theta \).
02

Simplify the equation

The expression \( \sin^2 \theta + 2 \sin \theta + 1 = 0 \) can be simplified. Notice that it follows the perfect square format: \((\sin \theta + 1)^2 = 0\).
03

Solve the simplified equation

To solve \((\sin \theta + 1)^2 = 0\), set the inner expression to zero: \(\sin \theta + 1 = 0\). Thus, we solve for \(\sin \theta\) by isolating it, giving \(\sin \theta = -1\).
04

Determine the general solutions

The equation \(\sin \theta = -1\) corresponds to \(\theta = \frac{3\pi}{2}\) within the interval \([0, 2\pi)\). \(\sin \theta = -1\) is only satisfied by this value within the given range.

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Key Concepts

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

Quadratic Equations
Quadratic equations are fundamental in mathematics, characterized by the standard form of \(ax^2 + bx + c = 0\). In our problem, we encounter a quadratic equation where \(x\) is replaced by \(\sin \theta\). This transformation helps us identify and solve trigonometric equations using similar techniques to those used in algebraic contexts.

The equation \(\sin^2 \theta + 2 \sin \theta + 1 = 0\) can be compared to the quadratic formula. Here:
  • \(a = 1\)
  • \(b = 2\)
  • \(c = 1\)
This allows us to transform and simplify the equation. Recognizing patterns, such as perfect squares, is key in finding solutions efficiently.

In this case, the equation \((\sin \theta + 1)^2 = 0\) is a perfect square, which simplifies the task as it mandates only a single solution is possible. Recognizing this form makes the process easier, providing the clear path to the solution that \(\sin \theta = -1\).
Unit Circle
The unit circle is a vital concept for understanding trigonometric equations. It is a circle with a radius of 1, centered at the origin of the coordinate system. Each point on the unit circle represents an angle measured from the positive x-axis, with coordinates \((\cos \theta, \sin \theta)\).

The significance of the unit circle becomes apparent when solving the equation \(\sin \theta = -1\). Around the unit circle, the sine function reaches its minimum value of \(-1\) at only one point within \(0 \leq \theta < 2\pi\), which is \(\theta = \frac{3\pi}{2}\).
  • \(\theta = \frac{3\pi}{2}\) represents the point \((0, -1)\) on the unit circle.
Therefore, knowledge of the unit circle helps us identify exact angle solutions quickly and accurately. It is essential in navigating through trigonometric identities and understanding their geometric interpretations.
Trigonometric Functions
Trigonometric functions encapsulate the relationships between angles and the lengths of triangles' sides. The most common trigonometric functions are sine, cosine, and tangent. In this exercise, the spotlight is on the sine function, \(\sin \theta\).

The sine function is defined as the ratio of the opposite side to the hypotenuse in a right-angled triangle. On the unit circle, \(\sin \theta\) is the y-coordinate for the corresponding angle \(\theta\). This insight is instrumental when solving trigonometric equations involving sine.

In the equation \(\sin \theta = -1\), understanding that this implies a specific y-coordinate aids us in quickly locating the relevant angle on the unit circle. The sine function’s periodic nature and its range from -1 to 1 are also crucial, as they define the possible outputs for \(\sin \theta\). Understanding these properties allows us to pinpoint all potential solutions, such as \(\theta = \frac{3\pi}{2}\) in this scenario.

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Most popular questions from this chapter

Although in general the statement \(\sin (A-B)=\sin A-\sin B\) is not true, it is true for some values. Determine some values of \(A\) and \(B\) that make this statement true.

Find the smallest positive values of \(x\) that make the statement true. Give the answer in degrees and round to two decimal places. $$e^{x}-\tan x=0$$

Solve the trigonometric equations exactly on the indicated interval, \(0 \leq x<2 \pi\). $$\sin (2 x)=\sqrt{3} \sin x$$

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Explain the mistake that is made. $$\text { Solve } \sqrt{2+\sin \theta}=\sin \theta \text { on } 0 \leq \theta \leq 2 \pi$$ Solution: Square both sides. \(2+\sin \theta=\sin ^{2} \theta\). Gather all terms to one side. \(\quad \sin ^{2} \theta-\sin \theta-2=0\). Factor. \(\quad(\sin \theta-2)(\sin \theta+1)=0\). Set each factor equal to zero. \(\sin \theta-2=0\) or \(\sin \theta+1=0\). Solve for \(\sin \theta . \quad \sin \theta=2 \quad\) or \(\quad \sin \theta=-1\). Solve \(\sin \theta=2\) for \(\theta . \quad\) no solution. Solve \(\sin \theta=-1\) for \(\theta . \quad \theta=\frac{3 \pi}{2}\). This is incorrect. What mistake was made?

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