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Chapter 6: Problem 10

How many solutions does the equation \((7 \sin x+2)(3 \cos x+3)\left(\tan ^{2} x-2\right)=0\) have over the interval \(0^{\circ}

Short Answer

Expert verified
There are 8 solutions over the interval \( 0^\circ < x \leq 360^\circ \).

Step by step solution

01

Understand the Equation

The given equation is \( (7 \sin x + 2)(3 \cos x + 3)(\tan ^{2} x - 2) = 0 \). An equation is zero if any of its factors are zero.
02

Solve for \( 7 \sin x + 2 = 0 \)

Set the factor equal to zero: \( 7 \sin x + 2 = 0 \). Solve for \sin x \:begin{align*}7 \sin x & = -2 \ \sin x & = \frac{-2}{7}end{align*}\find the values of x within the given interval for which \( \sin x = -\frac{2}{7} \).
03

Solve for \( 3 \cos x + 3 = 0 \)

Set the factor equal to zero: \( 3 \cos x + 3 = 0 \). Solve for \cos x \:begin{align*}3 \cos x & = -3 \ \cos x & = -1end{align*}\find the values of x within the given interval for which \( \cos x = -1 \).
04

Solve for \( \tan^2 x - 2 = 0 \)

Set the factor equal to zero: \( \tan^2 x - 2 = 0 \). Solve for \tan x \:begin{align*}\tan^2 x & = 2 \ \tan x & = \pm\sqrt{2}end{align*}\find the values of x within the given interval for which \( \tan x = \pm\sqrt{2} \).
05

Combine All Solutions

Combine the solutions from steps 2, 3, and 4. Ensure that all solutions are within the interval \( 0^\circ < x \leq \ 360^\circ \). Count the total number of solutions.

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Key Concepts

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

trigonometric functions
To solve any trigonometric equation, it's crucial to understand the main trigonometric functions: sine \( \sin(x) \), cosine \( \cos(x) \), and tangent \( \tan(x) \). These functions help us relate the angles of a right triangle to the lengths of its sides.
  • Sine \( \sin(x) \) is the ratio of the opposite side to the hypotenuse.
  • Cosine \( \cos(x) \) is the ratio of the adjacent side to the hypotenuse.
  • Tangent \( \tan(x) \) is the ratio of the opposite side to the adjacent side.

Understanding these ratios allows us to solve for angles when given a particular function's value. In our problem, we deal with different cases of \( \sin(x) \), \( \cos(x) \), and \( \tan(x) \). Solving each case independently and combining the results is key to finding all possible solutions.
solution intervals
When solving trigonometric equations, it's vital to consider the interval within which the solutions must lie. This means given a trigonometric equation, we have to find all the angles \( \ x \) that satisfy the equation within the specified range.
The interval for our problem is \( \ 0^{\circ} < x \leq 360^{\circ} \). This ensures we only consider angles within one complete rotation around the unit circle.
  • For \( 0^{\circ} \leq x < 360^{\circ} \), the circle begins at the positive x-axis and traces a full circle counterclockwise back to the same point.
  • Each trigonometric function repeats its values - a concept known as the period. Thus, for sine and cosine, the values start repeating every \( 360^{\circ} \).

Using this interval, we can find exact angular solutions for the factors in the equation, ensuring we stay within the proper range.
factorization of equations
Factorization simplifies complex trigonometric equations by breaking them into manageable parts. This allows us to solve for each part individually and then combine the solutions. For the given equation (7 \sin x +2)(3 \cos x + 3)(\tan^2 x - 2)=0, we need to treat each factor separately:
  • \(7 \sin x + 2 = 0 \) leads to \( \sin x = -\frac{2}{7}\),
  • \(3 \cos x + 3 =0 \) simplifies to \( \ cos x = -1 \),
  • \( \tan^{2}x - 2 = 0 \) becomes \( \tan x = \pm \sqrt{2} \)

Each of these equations can be solved within the interval \( 0^{\circ} < x \leq 360^{\circ} \). Once you solve each equation, you can combine all these solutions to determine the number of solutions in total.
trigonometric identities
Trigonometric identities are standard equations involving trigonometric functions true for all values of the variables. These identities help us manipulate and solve trigonometric equations. For instance, knowing that \( \tan(x) = \frac{\sin(x)}{\cos(x)} \) is an identity can simplify solving equations involving tangent.

Key identities include:
  • Pythagorean identities: \sin^2(x) + \cos^2(x) = 1
  • Angle sum and difference identities: \sin(A \pm B) = \sin(A)\cos(B) \pm \cos(A) \sin(B), \ \cos(A \pm B) = \cos(A)\cos(B) \mp \sin(A) \sin(B)
  • Double angle formulas: \sin(2x) = 2\sin(x) \cos(x), \ \cos(2x) = \cos^2(x) - \sin^2(x)

Applying these identities helps verify our solutions and can sometimes reveal hidden solutions or simplify complex expressions.

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Most popular questions from this chapter

When a ray of light hits a lens at angle of incidence \(\theta_{i},\) some of the light is refracted (bent) as it passes through the lens, and some is reflected by the lens. In the diagram, \(\theta_{r}\) is the angle of reflection and \(\theta_{t}\) is the angle of refraction. Fresnel equations describe the behaviour of light in this situation. a) Snells's law states that \(n_{1} \sin \theta_{i}=n_{2} \sin \theta_{\ell}\) where \(n_{1}\) and \(n_{2}\) are the refractive indices of the mediums. Isolate \(\sin \theta_{t}\) in this equation. b) Under certain conditions, a Fresnel equation to find the fraction, \(R,\) of light reflected is \(R=\left(\frac{n_{1} \cos \theta_{i}-n_{2} \cos \theta_{t}}{n_{1} \cos \theta_{i}+n_{2} \cos \theta_{t}}\right)^{2}\) Use identities to prove that this can be written as \(R=\left(\frac{n_{1} \cos \theta_{i}-n_{2} \sqrt{1-\sin ^{2} \theta_{t}}}{n_{1} \cos \theta_{i}+n_{2} \sqrt{1-\sin ^{2} \theta_{t}}}\right)^{2}\) c) Use your work from part a) to prove that $$ \begin{array}{l} \left(\frac{n_{1} \cos \theta_{i}-n_{2} \sqrt{1-\sin ^{2} \theta_{t}}}{n_{1} \cos \theta_{i}+n_{2} \sqrt{1-\sin ^{2} \theta_{t}}}\right)^{2} \\ =\left(\frac{n_{1} \cos \theta_{i}-n_{2} \sqrt{1-\left(\frac{n_{1}}{n_{2}}\right)^{2} \sin ^{2} \theta_{i}}}{n_{1} \cos \theta_{i}+n_{2} \sqrt{1-\left(\frac{n_{1}}{n_{2}}\right)^{2} \sin ^{2} \theta_{i}}}\right)^{2} \end{array} $$

Simplify and then give an exact value for each expression. a) \(\cos 40^{\circ} \cos 20^{\circ}-\sin 40^{\circ} \sin 20^{\circ}\) b) \(\sin 20^{\circ} \cos 25^{\circ}+\cos 20^{\circ} \sin 25^{\circ}\) c) \(\cos ^{2} \frac{\pi}{6}-\sin ^{2} \frac{\pi}{6}\) d) \(\cos \frac{\pi}{2} \cos \frac{\pi}{3}-\sin \frac{\pi}{2} \sin \frac{\pi}{3}\)

Rewrite each equation in terms of sine only. Then, solve algebraically for \(0 \leq x<2 \pi\). a) \(\cos 2 x-3 \sin x=2\) b) \(2 \cos ^{2} x-3 \sin x-3=0\) c) \(3 \csc x-\sin x=2\) d) \(\tan ^{2} x+2=0\)

Write each expression as a single trigonometric function. a) \(2 \sin \frac{\pi}{4} \cos \frac{\pi}{4}\) b) \(\left(6 \cos ^{2} 24^{\circ}-6 \sin ^{2} 24^{\circ}\right) \tan 48^{\circ}\) c) \(\frac{2 \tan 76^{\circ}}{1-\tan ^{2} 76^{\circ}}\) d) \(2 \cos ^{2} \frac{\pi}{6}-1\) e) \(1-2 \cos ^{2} \frac{\pi}{12}\)

When a polarizing lens is rotated through an angle \(\theta\) over a second lens, the amount of light passing through both lenses decreases by \(1-\sin ^{2} \theta\) a) Determine an equivalent expression for this decrease using only cosine. b) What fraction of light is lost when \(\theta=\frac{\pi}{6} ?\) c) What percent of light is lost when \(\theta=60^{\circ} ?\)
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