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Chapter 7: Problem 38

Solve each equation for solutions over the interval \(\left[0^{\circ}, 360^{\circ}\right) .\) Give solutions to the near. est tenth as appropriate. $$3 \cot ^{2} \theta-3 \cot \theta-1=0$$

Short Answer

Expert verified
The solutions are \( \theta = 38.7\degree, 218.7\degree, 104.5\degree, and 284.5\degree \).

Step by step solution

01

Recognize the quadratic form

The given equation is in the form of a quadratic equation in terms of \( \cot \theta \). Rewrite the equation as: \[ 3 \cot^2 \theta - 3 \cot \theta - 1 = 0 \]
02

Substitute variable

Let \( x = \cot \theta \) which transforms the equation into: \[ 3x^2 - 3x - 1 = 0 \]
03

Solve for variable using quadratic formula

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 3 \), \( b = -3 \), and \( c = -1 \). Substitute the values into the formula: \[ x = \frac{3 \pm \sqrt{9 + 12}}{6} = \frac{3 \pm \sqrt{21}}{6} = \frac{3 \pm 4.5826}{6} \]
04

Calculate the solutions

Calculate the two possible values for \( x \): \[ x_1 = \frac{3 + 4.5826}{6} = 1.264 \] and \[ x_2 = \frac{3 - 4.5826}{6} = -0.264 \]
05

Recover \( \theta \) from \( x \)

Recall that \( x = \cot \theta \), then we have \[ \cot \theta = 1.264 \] and \[ \cot \theta = -0.264 \]. Using the definition of \( \cot \theta \), find the corresponding \( \theta \) values: \[ \theta_1 = \arccot(1.264) = 38.7\degree \] and \[ \theta_2 = 180\degree - 38.7\degree = 180\degree + 180\degree - 38.7\degree = 218.7\degree \]. For \[ \cot \theta = -0.264 \]: \[ \theta_3 = \arccot(-0.264) = 104.5\degree \] and \[ \theta_4 = 180\degree + 104.5\degree = 284.5\degree \]

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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 polynomial equations of the form: \[ ax^2 + bx + c = 0 \] ahol \( a \), \( b \) 茅s \( c \) konstansok. Ezek az egyenletek fontosak a matematika sok 谩g谩ban, ez茅rt, a meg茅rt茅s眉k kulcsfontoss谩g煤. Az egyik f艖 m贸dszer a megold谩sukhoz a kvadratikus k茅plet haszn谩lata: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. Ennek haszn谩lata megk枚veteli, hogy meghat谩rozzuk az \( a \), \( b \), 茅s \( c \) 茅rt茅keit, majd helyettes铆ts眉k ezeket az egyenletbe. Ebben a feladatban az \( x \) 茅rt茅keket \( cot \theta \) -vel helyettes铆tj眉k, 铆gy a kapott 茅rt茅kek \( \theta \) -hez rendelhet艖ek.
Cotangent Function
The cotangent function, denoted as cot, is the reciprocal of the tangent function. It is defined as: \[ \text{cot} \theta = \frac{\text{cos} \theta}{\text{sin} \theta} \]. This function is significant in solving trigonometric equations as it relates the angles of a triangle to the lengths of its sides. For solving equations, it helps to convert cotangent values to corresponding angles using its inverse function. When \( \text{cot} \theta \) is positive, \( \theta \) lies in the first or third quadrant. When \( \text{cot} \theta \) is negative, \( \theta \) lies in the second or fourth quadrant.
Inverse Trigonometric Functions
Inverse trigonometric functions are essential for finding angles when trigonometric values are known. The inverse of the cotangent function is the arccotangent. For example, if \( \text{cot} \theta = 1.264 \), then \( \theta = \text{arccot}(1.264) \), producing an angle. These functions help translate between trigonometric ratios and angles. They can also extend solutions within one period of the trigonometric function, usually adding multiples of the period. For angles between \( 0^\text{掳} \)and \( 360^\text{掳} \), this helps find all relevant solutions.

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Most popular questions from this chapter

Solve each problem. When a musical instrument creates a tone of \(110 \mathrm{Hz}\). it also creates tones at \(220,330,440,550,660, \ldots\) Hz. A small speaker cannot reproduce the \(110-\mathrm{Hz}\) vibration but it can reproduce the higher frequencies, which are the upper harmonics. The low tones can still be heard because the speaker produces difference tones of the upper harmonics. The difference between consecutive frequencies is \(110 \mathrm{Hz}\), and this difference tone will be heard by a listener. (Source: Benade, A.. Fundamentals of Musical Acoustics, Dover Publications.) (a) We can model this phenomenon using a graphing calculator. In the window \([0,0.03]\) by \([-1,1],\) graph the upper harmonics represented by the pressure $$ P=\frac{1}{2} \sin [2 \pi(220) t]+\frac{1}{3} \sin [2 \pi(330) t]+\frac{1}{4} \sin [2 \pi(440) t] $$ (b) Estimate all \(t\) -coordinates where \(P\) is maximum. (c) What does a person hear in addition to the frequencies of \(220,330,\) and \(440 \mathrm{Hz} ?\) (d) Graph the pressure produced by a speaker that can vibrate at \(110 \mathrm{Hz}\) and above.

Verify that each equation is an identity. $$\tan 2 \theta=\frac{-2 \tan \theta}{\sec ^{2} \theta-2}$$

Use a calculator to find each value. Give answers as real numbers. $$\cot (\arccos 0.58236841)$$

Alternating Electric Current In the study of alternating electric current, instantaneous voltage is modeled by E=E_{\max } \sin 2 \pi f t where \(f\) is the number of cycles per second, \(E_{\max }\) is the maximum voltage, and \(t\) is time in seconds. (a) Solve the equation for \(t\) (b) Find the least positive value of \(t\) if \(E_{\max }=12, E=5,\) and \(f=100 .\) Use a calculator.

Verify that each trigonometric equation is an identity. $$\frac{1-\cos x}{1+\cos x}=\csc ^{2} x-2 \csc x \cot x+\cot ^{2} x$$
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