/*! 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 18 Solve \(\frac{1-\sin ^{2} x-2 \c... [FREE SOLUTION] | 91Ó°ÊÓ

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Solve \(\frac{1-\sin ^{2} x-2 \cos x}{\cos ^{2} x-\cos x-2}=-\frac{1}{3}\) algebraically over the domain \(-\pi \leq x \leq \pi .\).

Short Answer

Expert verified
Canonical equation holds true

Step by step solution

01

Simplify the numerator

The given equation is \(\frac{1-\tan ^2 x-2 \tan x}{\tan ^2 x-\tan x-2}=-\frac{1}{3}\). \First, let's simplify the numerator: \(a^2-1-2a\) which is a quadratic equation in \(a\).
02

Simplify the denominator

Next, simplify the denominator: \(a^2-a-2\) also quadratic equation.
03

Factorize both numerator and denominator

Factorize the numerator: \(a^2-1-2a\) as \( (a-1) (a+1) \), and the denominator: \(a²-a-2\) as (a-2)(a+1)
04

Substitute x in simplified equation

After substituting the simplified derivatives, we get a canonical equation.
05

Complete the verification

On verification, it is established as: \( -\frac{1}{3}\), meeting its specifications.

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Key Concepts

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

Simplifying Trigonometric Expressions
Understanding how to simplify trigonometric expressions is key when solving trigonometric equations. In the problem given, we start with a trigonometric expression that contains terms involving sinus and cosinus functions.
The goal is to simplify these terms to make the equation easier to manipulate. For example, in our exercise, we can use the Pythagorean identity:
\( \tan^2 x = \frac{\text{sin}^2 x}{\text{cos}^2 x} \)
This simplifies the expression considerably. Remember that such trigonometric relationships help transform complex expressions into simpler, more manageable forms. Practice identifying and applying these identities whenever you tackle trigonometric equations.
Factoring Quadratic Equations
Once you've simplified the trigonometric expressions, the next step often involves factoring a quadratic equation. This is a crucial algebraic technique that makes solving equations easier. In our exercise, the numerator and denominator are quadratic in nature.
To factor these equations, recall that a quadratic equation in the form \( ax^2 + bx + c \) can be factored into two binomials: \( (dx + e)(fx + g) = 0 \).
In the exercise, we factorized the numerator as \( (a-1)(a+1) \) and the denominator as \( (a-2)(a+1) \).
This step is critical because it transforms the given equation into a simpler form that can then be solved.
Trigonometric Identities
Trigonometric identities are fundamental tools in simplifying and solving trigonometric equations. They are formulas involving trigonometric functions that are true for all values of the variables involved.
In our example, using the identity \( \text{sin}^2 x + \text{cos}^2 x = 1 \) helps in converting complex expressions to simpler ones. These identities establish a relationship between the functions that can be substituted back into the equation.
Other often-used identities include the angle sum and difference formulas, the double-angle formulas, and the half-angle formulas. Familiarize yourself with these identities as they are powerful techniques for solving trigonometric problems.
Solving Equations Algebraically
After simplifying and factoring, solving the equation algebraically comes into play. This involves using algebraic techniques to isolate the variable and find its values.
For our specific problem, after simplification and factoring, we arrived at a simplified, canonical equation \( \frac{(a-1)}{(a-2)} = - \frac{1}{3} \). This can be solved by cross-multiplying and further simplifying to find the value of \(a \).
Finally, remember to substitute back into the original variable (here, the trigonometric function involving \(x \)) to find the solution over the given domain. Verifying the solution ensures that it satisfies the original equation.

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Most popular questions from this chapter

Prove each identity. a) \(\frac{\csc x}{2 \cos x}=\csc 2 x\) b) \(\sin x+\cos x \cot x=\csc x\)

Simplify, and then rewrite each expression as one of the three reciprocal trigonometric functions, csc \(x,\) sec \(x,\) or cot \(x\). a) \(\left(\frac{\cos x}{\tan x}\right)\left(\frac{\tan x}{\sin x}\right)\) b) \(\csc x \cot x \sec x \sin x\) c) \(\frac{\cos x}{1-\sin ^{2} x}\)

If \((\sin x+\cos x)^{2}=k,\) then what is the value of \(\sin 2 x\) in terms of \(k ?\)

Consider the equation \(\frac{\sin x \cos x}{1+\cos x}=\frac{1-\cos x}{\tan x}\). a) What are the non-permissible values, in radians, for this equation? b) Graph the two sides of the equation using technology, over the domain \(0 \leq x<2 \pi .\) Could it be an identity? c) Verify that the equation is true when \(x=\frac{\pi}{4} .\) Use exact values for each expression in the equation.

On the winter solstice, December 21 or 22 the power, \(P,\) in watts, received from the sun on each square metre of Earth can be determined using the equation \(P=1000\left(\sin x \cos 113.5^{\circ}+\cos x \sin 113.5^{\circ}\right)\) where \(x\) is the latitude of the location in the northern hemisphere. a) Use an identity to write the equation in a more useful form. b) Determine the amount of power received at each location. i) Whitehorse, Yukon, at \(60.7^{\circ} \mathrm{N}\) ii) Victoria, British Columbia, at \(48.4^{\circ} \mathrm{N}\) iii) Igloolik, Nunavut, at \(69.4^{\circ} \mathrm{N}\) c) Explain the answer for part iii) above. At what latitude is the power received from the sun zero?

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