Problem 31 - A Graphical Approach to Precalculus with Limits

Question

Solve \((\mathbf{a}) f(x)=0,(\mathbf{b}) f(x)>0,\) and \((\mathbf{c}) f(x)<0\) over the interval \([0,2 \pi)\) $$f(x)=\sin ^{2} x \cos x-\cos x$$

Step-by-Step Solution

Answer

(a) \(x=\frac{\pi}{2}, \frac{3\pi}{2}\); (b) \(x\in(\frac{\pi}{2}, \frac{3\pi}{2})\); (c) \(x\in[0,\frac{\pi}{2})\cup(\frac{3\pi}{2},2\pi)\)."

Step 1: Factor the Expression

We start by factoring the expression given for \(f(x)\). The expression is \(f(x) = \sin^2 x \cos x - \cos x\). We notice that \(\cos x\) is a common factor, thus:\[f(x) = \cos x (\sin^2 x - 1)\]

Step 2: Recognize Trigonometric Identity

Recognize that \(\sin^2 x - 1\) can be rewritten using the Pythagorean identity \(\sin^2 x = 1 - \cos^2 x\). Hence,\[\sin^2 x - 1 = -(\cos^2 x - 1) = -(1 - \cos^2 x)\] simplifies to \(-\cos^2 x\). Thus, we rewrite our function:\[f(x) = \cos x (-\cos^2 x) = -\cos^3 x\]

Step 3: Solve for (a) \(f(x)=0\)

Set \(-\cos^3 x = 0\). This gives us \(\cos x = 0\). We solve for \(x\) over the interval \([0, 2\pi)\).\[\cos x = 0 \Rightarrow x = \frac{\pi}{2}, \frac{3\pi}{2}\]

Step 4: Solve for (b) \(f(x)>0\)

Consider where \(-\cos^3 x > 0\). This inequality holds when \(\cos^3 x < 0\), which occurs when \(\cos x < 0\).Find \(x\) where \(\cos x < 0\) within \([0, 2\pi)\):- The cosine function is negative in the second and third quadrants: - Solutions in the interval are: - From \(\pi/2\) to \(3\pi/2\)So, \(x \in \left(\frac{\pi}{2}, \frac{3\pi}{2}\right)\).

Step 5: Solve for (c) \(f(x)<0\)

Now consider where \(-\cos^3 x < 0\). This inequality holds true when \(\cos^3 x > 0\), which occurs when \(\cos x > 0\).Find \(x\) where \(\cos x > 0\) within \([0, 2\pi)\):- The cosine function is positive in the first and fourth quadrants: - Solutions in the interval are: - From \(0\) to \(\pi/2\) and from \(3\pi/2\) to \(2\pi\)So, \(x \in [0, \frac{\pi}{2}) \cup (\frac{3\pi}{2}, 2\pi)\).

Key Concepts

Factoring Expressions

When dealing with algebraic or trigonometric expressions, factoring is often one of the first approaches to simplify the problem. It involves breaking down a complex expression into simpler components or factors.
In the given function, \( f(x) = \sin^2 x \cos x - \cos x \), the common factor is \( \cos x \). Factoring it out, we get:

• \( f(x) = \cos x (\sin^2 x - 1) \)

Factoring simplifies the problem by reducing the number of trigonometric terms we need to consider. It helps us easily identify and apply trigonometric identities, which is our next step.

Trigonometric Identities

Trigonometric identities are equations that relate various trigonometric functions to one another. They are crucial when simplifying expressions or proving other trigonometric equations. The Pythagorean identity is especially useful in this exercise:

• \( \sin^2 x = 1 - \cos^2 x \)

Using this identity, the expression \( \sin^2 x - 1 \) transforms into a simpler form:

• \( \sin^2 x - 1 = -(1 - \cos^2 x) = -\cos^2 x \)

Thus, the original function becomes:

• \( f(x) = \cos x (-\cos^2 x) = -\cos^3 x \)

Understanding and applying trigonometric identities streamline the process of solving equations by reducing the complexity of expressions.

Inequalities

Inequalities in trigonometry involve understanding when certain conditions about an expression hold true.
For the function \( f(x) = -\cos^3 x \), we solve three types of inequalities:

• When \( f(x) = 0 \), or equivalently \( -\cos^3 x = 0 \), it implies \( \cos x = 0 \).
• When \( f(x) > 0 \), meaning \( -\cos^3 x > 0 \), it translates to \( \cos^3 x < 0 \), so \( \cos x < 0 \).
• When \( f(x) < 0 \), which is \( -\cos^3 x < 0 \), it indicates \( \cos^3 x > 0 \), so \( \cos x > 0 \).

Each inequality guides us to solutions in different intervals based on the behavior of the cosine function over \([0, 2\pi)\). Understanding these conditions helps us find solutions within specified ranges.

Interval Notation

Interval notation is a concise way of writing subsets of the real number line. It is a handy tool in mathematics to indicate the set of solutions or domain of a function. Here’s how interval notation works in this context:

• The interval \( [0, 2\pi) \) specifies the domain we are interested in, namely from \( 0 \) (included) to \( 2\pi \) (excluded).
• For \( f(x) = 0 \), solutions are at the boundaries where \( \cos x = 0 \), which are \( x = \frac{\pi}{2}, \frac{3\pi}{2} \).
• For \( f(x) > 0 \), the solutions where \( \cos x < 0 \) occur in the interval \( (\frac{\pi}{2}, \frac{3\pi}{2}) \).
• When \( f(x) < 0 \), \( \cos x > 0 \), the intervals are \( [0, \frac{\pi}{2}) \cup (\frac{3\pi}{2}, 2\pi) \).

Interval notation provides an organized approach to express solution sets and ensures clarity when describing continuous subsets of numbers.