Problem 125 - Algebra and Trigonometry

Question

solve each equation on the interval \([0,2 \pi) .\) \(2 \cos ^{3} x+\cos ^{2} x-2 \cos x-1=0\) (Hint: Use factoring by grouping.)

Step-by-Step Solution

Answer

The solutions on the interval \(0,2 \pi\) are \(x=0, \pi, 2\pi, 2\pi/3, \)and \(4\pi/3\).

Step 1: Rewrite the Equation

Group the terms in the equation so you can see if the equation factors by grouping. Here, rearrange the terms in pairs that have common factors: \((2 \cos ^{3} x + \cos ^{2} x)- (2 \cos x + 1)=0\)

Step 2: Factor by Grouping

Factor out an \(\cos ^{2} x \) in the first group and a \(-1\) in the second, and you see that you are left with the same expression in brackets:\(\cos ^{2} x (2 \cos x +1) -1(2 \cos x +1)=0\)

Step 3: Continue Factoring

Then we see an expression of the form \(a(b)-c(b)\) which is equivalent to \((a-c)(b)\). Thus, factor out \((2 \cos x +1)\), we should get:\((\cos ^{2} x -1) (2 \cos x +1 ) = 0\)

Step 4: Apply Cosine Function Properties

The final step in factoring recognizes that \(\cos ^{2} x - 1\) is equivalent to \(-\sin ^{2} x\). Therefore, we have:\(-\sin ^{2} x(2 \cos x +1 ) = 0\)

Step 5: Set Each Factor Equal to Zero

Finally, for a product to be zero, at least one of the factors must be zero:\(-\sin ^{2} x = 0\) and \(2 \cos x + 1 = 0\)

Step 6: Find the Possible Values of x

Solving for \(x\) gives:For \(\sin ^{2} x = 0\), \(x=0, \pi, 2\pi\)and,For \(2 \cos x + 1 = 0\), \(\cos x=-0.5\) which gives \(x = 2\pi/3, 4\pi/3\)

Key Concepts

Factoring by grouping

Factoring by grouping is a useful technique for solving polynomial equations, including trigonometric equations. The idea behind this method is to rearrange and group terms in a way that reveals common factors.

In our equation, \(2 \cos^{3} x + \cos^{2} x - 2 \cos x - 1 = 0\), we begin by clustering similar terms:

• First Group: \(2 \cos^{3} x + \cos^{2} x\)
• Second Group: \(-2 \cos x - 1\)

This rearrangement allows us to factor each group independently, pulling out common factors.

In the first group, \(\cos^{2} x\) is common; in the second group, \(-1\) is the factor to extract.
This leads us to write the expression as \((2 \cos x + 1)\) for both groups. By doing this, we are set to identify a common factor: \((\cos^{2} x - 1)(2 \cos x + 1) = 0\).
This method simplifies complex polynomials into manageable parts, revealing potential solutions at each step.

Cosine function properties

Understanding the cosine function is crucial for solving trigonometric equations. The cosine function, \(\cos x\), is periodic, continuous, and oscillates between -1 and 1.
It repeats its values every \(2\pi\) radians, making it essential to consider the interval of solutions, especially when solving equations over \([0, 2\pi)\).

• The squared term \(\cos^{2} x\) becomes useful in identity transforms, as we saw with \(\cos^{2} x - 1\).
• This expression is a recalled identity: \(\cos^{2} x - 1 = -\sin^{2} x\), which allows transformation into another trigonometric function.

Recognizing these properties helps simplify equations and find solutions within the specified interval.

The cosine function's behavior at multiples of \(\pi\) and solutions where the cosine equals specific values, such as \(0\) or \(-0.5\), provide key points.
These key points facilitate identification of where the equation equals zero, leading toward solutions satisfying the original equation.

Solving trigonometric equations

Solving trigonometric equations involves isolating the trigonometric part and finding specific solutions within a given interval. After factoring the given equation into \((\cos^{2} x - 1)(2 \cos x + 1) = 0\), it's necessary to explore each factor individually.

• For the first factor, \(-\sin^{2} x = 0\), using the identity transforms, you rise to \(\sin x = 0\), which is solved within the interval \([0, 2\pi)\).
• For the second factor, \(2 \cos x + 1 = 0\), solve \(\cos x = -0.5\).

These individual solutions exist where the cosine or sine functions reach respective values that zero out each factor.

Specifically, solving \(\sin x = 0\) brings solutions \(x=0, \pi, 2\pi\). Here, \(2\pi\) is singular because it's not in the open interval \([0, 2\pi)\).
Similarly, \(\cos x = -0.5\) yields angles at \(x = 2\pi/3, 4\pi/3\) derived from reference angles across quadrants.

Listing these solutions carefully ensures that each angle fits within the round closed interval, providing proper satisfaction to the initial trigonometric equation.