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Solve each equation for solutions over the interval \([0,2 \pi)\) by first solving for the trigonometric finction. Do not use a calculator. $$(\csc x+2)(\csc x-\sqrt{2})=0$$

Short Answer

Expert verified
The solutions are \(x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{7\pi}{6}, \frac{11\pi}{6}\).

Step by step solution

01

Identify Zero Products

The given equation is a product of two factors: \((\csc x + 2)\) and \((\csc x - \sqrt{2})\). According to the zero-product property, if the product of two expressions is zero, then at least one of the expressions must be zero. Thus, we set each factor equal to zero separately: \(\csc x + 2 = 0\) and \(\csc x - \sqrt{2} = 0\).
02

Solve the First Equation

Solve \(\csc x + 2 = 0\):\[ \begin{align*} \csc x + 2 &= 0 \ \csc x &= -2. \end{align*}\]The cosecant function is the reciprocal of the sine function, so \(\csc x = \frac{1}{\sin x}\). Thus, \(\sin x = -\frac{1}{2}\).
03

Find Angles for First Equation

Determine angles \(x\) for which \(\sin x = -\frac{1}{2}\). Over the interval \([0, 2\pi)\), these angles are \(x = \frac{7\pi}{6}\) and \(x = \frac{11\pi}{6}\). These correspond to the points where sine is -0.5, occurring in the third and fourth quadrants.
04

Solve the Second Equation

Solve \(\csc x - \sqrt{2} = 0\):\[ \begin{align*} \csc x - \sqrt{2} &= 0 \ \csc x &= \sqrt{2}. \end{align*}\]Therefore, \(\sin x = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}\).
05

Find Angles for Second Equation

Find angles \(x\) for which \(\sin x = \frac{\sqrt{2}}{2}\). These are \(x = \frac{\pi}{4}\) and \(x = \frac{3\pi}{4}\), since the sine function has these values in the first and second quadrants.
06

Compile All Solutions

Collect all solutions from both cases: The solutions for \(\sin x = -\frac{1}{2}\) are \(x = \frac{7\pi}{6}\) and \(x = \frac{11\pi}{6}\). The solutions for \(\sin x = \frac{\sqrt{2}}{2}\) are \(x = \frac{\pi}{4}\) and \(x = \frac{3\pi}{4}\). Thus, the complete solution set over the interval \([0, 2\pi)\) is \(x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{7\pi}{6}, \frac{11\pi}{6}\).

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Key Concepts

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

Cosecant Function
The cosecant function, denoted as \(\csc x\), is one of the fundamental trigonometric functions and is defined as the reciprocal of the sine function. Mathematically, it is expressed as:\[ \csc x = \frac{1}{\sin x} \]This means that wherever the sine of an angle is zero, the cosecant function is undefined. This happens at specific points such as \(x = 0, \pi, 2\pi\), and so on.
  • **Range and Domain:** The range of \(\csc x\) is the set of all real numbers except for the open interval \((-1, 1)\). This is because the sine function has values that range from -1 to 1, except that \(\csc x\) is not defined at these values.
  • **Key Angles:** Often, when solving trigonometric equations involving cosecant, we look for angles where sine equals commonly known values like \(-\frac{1}{2}\), or \(\frac{\sqrt{2}}{2}\).
Understanding the cosecant function is critical in solving equations like the original exercise, where we had to find solutions for expressions involving \(\csc x\). By knowing the relationship between sine and cosecant, we can easily manipulate and solve such equations.
For example, if \(\csc x = -2\), this implies that \(\sin x = -\frac{1}{2}\). This reciprocal nature aids us in finding solutions over given intervals.
Zero-Product Property
The zero-product property is a fundamental algebraic principle stating that if the product of two or more factors is equal to zero, then at least one of the factors must be zero. This property underpins the solving of many algebraic equations and is equally applicable to trigonometric equations.
In our exercise, we used the zero-product property to solve:\[ (\csc x + 2)(\csc x - \sqrt{2}) = 0 \]By the zero-product property:
  • Set \(\csc x + 2 = 0\)
  • Set \(\csc x - \sqrt{2} = 0\)
Each equation can be solved independently to find the possible values of \(x\). The solutions to these individual equations comprise the complete solution to the original equation.
This approach is extremely useful because it simplifies complex equations into manageable parts, allowing us to find potential solutions more quickly.
Interval Notation
Interval notation is a method of denoting a range of numbers, useful in mathematics to describe solutions or values belonging to a particular set. In the context of trigonometric equations, interval notation helps specify over which values we are interested in finding solutions, such as \[ [0, 2\pi) \]Here's how to understand this interval:
  • The "[" symbol means that the starting point, 0, is included in the interval.
  • The ")" symbol means that the endpoint, \(2\pi\), is not included, indicating an open interval at \(2\pi\).
This often defines the domain over which we are examining our solutions. In trigonometry, solving within a specific interval is crucial as trigonometric functions are periodic and their solutions repeat over different intervals. Thus, specifying a particular interval, such as \([0, 2\pi)\), prevents contradiction with overlapping solutions from other periods.
Comprehending interval notation makes problem-solving in mathematics efficient and clear, ensuring we consider only the relevant solutions.

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Most popular questions from this chapter

Solve each equation ( \(x\) in radians and \(\theta\) in degrees) for all exact solutions where appropriate. Round approximale values in radians to four decimal places and approximate values in degrees to the nearest tenth. Write answers using the least possible nonnegative angle measures. $$-2 \cos 2 x=\sqrt{3}$$

Use a graphical method to solve each equation over the interval \([0,2 \pi) .\) Round values to the nearest thousandth. $$\sin \frac{x}{2}+\cos 3 x=0$$

Determining Wattage Amperage is a measure of the amount of electricity that is moving through a circuit, while voltage is a measure of the force pushing the electricity. The wattage \(W\) consumed by an electrical device can be determined by calculating the product of amperage \(I\) and voltage \(V .\) (Source: Wilcox, G. and C. Hesselberth, Electricity for Engineering Technology, Allyn \& Bacon.) (a) A household circuit has voltage $$V=163 \sin 120 \pi t$$ when an incandescent light bulb is turned on with amperage $$I=1.23 \sin 120 \pi t$$ Graph the wattage $$W=V I$$ that is consumed by the light bulb over the interval \(0 \leq t \leq 0.05\) (b) Determine the maximum and minimum wattages used by the light bulb. (c) Use identities to find values for \(a, c,\) and \(\omega\) so that $$ W=a \cos \omega t+c $$ (d) Check your answer in part (c) by graphing both expressions for \(W\) on the same coordinate axes. (e) Use the graph from part (a) to estimate the average watt. age used by the light. How many watts do you think this incandescent light bulb is rated for?

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Give the exact real number value of each expression. Do not use a calculator. $$\tan \left(\sin ^{-1} \frac{8}{17}+\tan ^{-1} \frac{4}{3}\right)$$

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