Question

The solution set of \((2 \cos x-1)(3+2 \cos x)=0\) in the interval \(0 \leq x \leq 2 \pi\) is (a) \(\left\\{\frac{\pi}{3}, \frac{5 \pi}{3}\right\\}\) (b) \(\left\\{\frac{\pi}{3}\right\\}\) (c) \(\left\\{\frac{\pi}{3}, \frac{5 \pi}{3}, \cos ^{-1}\left(-\frac{3}{2}\right)\right\\}\) (d) None of these

Short answer

Answer: \(\left\{\frac{\pi}{3}, \frac{5 \pi}{3}\right\}\)

Step-by-step solution

Solve inequalities individually

First, we solve each factor individually. So, we need to find the values of x for which: 1) \(2 \cos x -1 = 0\) 2) \(3 + 2 \cos x = 0\)

Solve Factor 1

For the first factor, we have: \(2 \cos x -1 = 0 \Rightarrow 2 \cos x = 1 \Rightarrow \cos x = \frac{1}{2}\) Now, we need to find the values of \(x\) such that the cosine function is equal to \(\frac{1}{2}\). It's known that: \( \cos \left( \frac{\pi}{3} \right) = \frac{1}{2} \) and \( \cos \left( \frac{5\pi}{3} \right) = \frac{1}{2} \) So, the values of \(x\) that satisfy the first factor are \(x_1 = \frac{\pi}{3}\) and \(x_2 = \frac{5\pi}{3}\).

Solve Factor 2

For the second factor, we have: \(3 + 2 \cos x = 0 \Rightarrow 2 \cos x = -3 \Rightarrow \cos x = -\frac{3}{2}\) However, we know that the range of cosine function is \(-1 \leq \cos x \leq 1\). Here, \(\cos x = -\frac{3}{2}\) is out of the range. Therefore, no value of \(x\) can satisfy this factor.

Determine the solution set

Based on Step 2 and Step 3, we find that the only solutions to the given equation are \(x = \frac{\pi}{3}\) and \(x = \frac{5\pi}{3}\), which are within the given interval \(0 \leq x \leq 2 \pi\). Comparing these solutions to the given options, they match with option (a): (a) \(\left\{\frac{\pi}{3}, \frac{5 \pi}{3}\right\}\)

Key concepts

Solution Set

The term **solution set** refers to all values of a variable that satisfy a given equation. In trigonometry, especially when working with angles, it becomes crucial to find which specific angles within a defined range satisfy the equation. For our given exercise, the equation \((2 \cos x -1)(3 + 2 \cos x) = 0\) provides two separate conditions, which are essentially two different trigonometric equations:

• \(2 \cos x - 1 = 0\)
• \(3 + 2 \cos x = 0\)

Solving each of these equations individually will give us potential solutions within the interval \([0, 2\pi]\). However, we must ensure these solutions are valid in context, particularly considering the typical range constraints of trigonometric functions like cosine.
Ultimately, the valid solution set was determined to be \(x = \frac{\pi}{3}\) and \(x = \frac{5\pi}{3}\), which align perfectly with option (a) in the exercise.

Cosine Function

The **cosine function** is one of the primary trigonometric functions, denoted as \(\cos\). It relates the adjacent side of a right triangle to its hypotenuse. The function is periodic, meaning it repeats its values in regular intervals, specifically every \(2\pi\).
For any angle \(x\), the cosine function returns a value within the range of - \([-1, 1]\).To solve the equation \((2 \cos x -1) = 0\) at the core of our problem, we rearrange it to find when the cosine equals \(\frac{1}{2}\). Knowing specific angle values where the cosine function gives \(\frac{1}{2}\) helps simplify this process. Specifically, knowing:

• \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\)
• \(\cos\left(\frac{5\pi}{3}\right) = \frac{1}{2}\)

In this context, the cosine's periodic nature ensures that we recognize other equivalent angles that also have the cosine value of \(\frac{1}{2}\), spread around its cycle.

Intervals in Trigonometry

In trigonometry, an **interval** usually refers to a specified range of angles measured in radians. Understanding these intervals is vital as trigonometric functions like cosine are periodic and repeat their values over fixed intervals. Our problem's interval is the standard range of one complete cycle of the trigonometric circle:- \([0, 2\pi]\)This interval signifies a full rotation of 360 degrees in radians, which includes every possible angle the equation might satisfy.
Within this interval, we methodically determine which angles produce the values required by the cosine function in our trigonometric equation. It is crucial not just to find values that theoretically satisfy the equation but also to confirm that they reside within the given boundary of \([0, 2\pi]\). Recognizing this constraint helps us eliminate any extraneous solutions.

Trigonometric Identities

**Trigonometric identities** are essential equations involving trigonometric functions that hold true for every angle. They simplify complex trigonometric expressions and help solve equations effectively. While solving the exercise equation \((2 \cos x-1)(3+2 \cos x) = 0\), it's essential to recognize that trigonometric identities can further break down how we handle these factors.
For instance, the identity:- \(\cos^2 x + \sin^2 x = 1\)can be vital for transforming cosine into sine if necessary, though it wasn't directly applicable here with our cosine-based approach.Another useful identity relates to the cosine double angle formula: - \(\cos 2x = 2\cos^2 x - 1\)Our task simplifies when focusing on only solving for factor-derived equations where the trigonometric identity isn't directly in use, but knowing these expressions exist broadens our toolkit for approaching complex problems in trigonometry. Understanding identities like the above can often reveal alternative pathways or validate derived solutions when cross-referencing results.