Problem 36 - Precalculus with Limits

Question

In Exercises 25-38, find all solutions of the equation in the interval \( [0, 2\pi) \). \( \cos x + \sin x \tan x = 2 \)

Step-by-Step Solution

Answer

The equation has no real solution within the interval \( [0, 2\pi) \).

Step 1: Express the equation in terms of sine and cosine

What needs to be done first is to convert the tangent function, which is actually the ratio of sine to cosine, i.e. \( \tan x = \frac{\sin x}{\cos x} \). Substituting this into the equation gives: \( \cos x + \sin x \cdot \frac{\sin x}{\cos x} = 2 \)

Step 2: Simplify the equation

By carrying out the multiplication, the equation becomes \( \cos x + \sin^2 x = 2 \). However, we know that \( \sin^2 x = 1 - \cos^2 x \) so we can further simplify this equation to \( \cos x + 1 - \cos^2 x = 2 \). Rearranging the order gives: \( \cos^2 x - \cos x - 1 = 0 \). This is a quadratic function, which we set to zero to find the values of \( \cos x \).

Step 3: Find the roots

The roots of the quadratic equation can be found using the quadratic formula: \( \cos x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). In our equation, \( a = 1 \), \( b = -1 \), and \( c = -1 \). Substituting these values into the formula gives: \( \cos x = \frac{1 \pm \sqrt{1 + 4}}{2} \), which simplifies to \( \cos x = \frac{1 \pm \sqrt{5}}{2} \). The roots are \( \frac{1 + \sqrt{5}}{2} \) and \( \frac{1 - \sqrt{5}}{2} \). However, the latter root is not a possible value for the cosine function in the interval [0, 2pi).

Step 4: Finds corresponding x

Using the inverse cosine, solve \( x = \arccos\left(\frac{1 + \sqrt{5}}{2}\right) \) to find the angle 'x'. However, this number is greater than 1 which is out of the range of \(\arccos(x)\), and therefore the equation has no real solution within the given interval.

Key Concepts

Cosine Function

The cosine function, represented as \( \cos x \), is one of the fundamental trigonometric functions. It associates each angle with the x-coordinate of a point on the unit circle. The cosine function has a range of values between -1 and 1. This means any output of the cosine function must lie within this interval.Because the cosine function deals with angles and their projections onto the x-axis, it's crucial to remember these key points:

• Cosine value of 0° (or 0 radians) is 1.
• Cosine value of 90° (or \( \frac{\pi}{2} \) radians) is 0.
• Cosine value of 180° (or \( \pi \) radians) is -1.
• Cosine value of 270° (or \( \frac{3\pi}{2} \) radians) is 0.

For angles outside the first quadrant, the cosine function continues its sinusoidal behavior: it repeats every \( 2\pi \) radians (360°), leading to both positive and negative values along the way.

Sine Function

Similar to the cosine function, the sine function \( \sin x \) represents one of the primary trigonometric functions. It relates angles to the y-coordinate of a point on the unit circle. Just like cosine, the sine function also has a range from -1 to 1.Some key properties of the sine function include:

• Sine value of 0° (or 0 radians) is 0.
• Sine value of 90° (or \( \frac{\pi}{2} \) radians) is 1.
• Sine value of 180° (or \( \pi \) radians) is 0.
• Sine value of 270° (or \( \frac{3\pi}{2} \) radians) is -1.

The sine function, like the cosine, follows a repeating pattern every \( 2\pi \) as it moves around the circle. Thus, as you consider solutions involving sine or cosine, their periodic nature and defined range (-1 to 1) must always be kept in mind.

Quadratic Formula

The quadratic formula is a powerful tool for finding roots of quadratic equations. A quadratic equation is generally given by the form \( ax^2 + bx + c = 0 \). The formula to solve for \( x \) is:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Using this formula involves these steps:

• Identify coefficients \( a \), \( b \), and \( c \) from the quadratic equation.
• Calculate the discriminant \( b^2 - 4ac \).
• If the discriminant is positive, there are two different real roots; if zero, one real root; if negative, no real roots (complex roots).
• Substitute the coefficients and discriminant into the formula to find the roots.

The quadratic formula not only provides us solutions but also helps in identifying the characteristics of the roots based on the discriminant value. In a trigonometric context like here, the values must be checked against suitable function ranges, such as that of cosine and sine.

Interval Notation

Interval notation is a mathematical shorthand for representing intervals on the real number line. In solving trigonometric equations, we frequently use interval notation to denote the set of possible solutions within certain boundaries. For example, the interval \( [0, 2\pi) \) describes all angles from 0 to \( 2\pi \) radians, including 0 but not \( 2\pi \).Important aspects of interval notation are:

• "[" or "]" brackets mean the endpoint is included (closed interval).
• "(" or ")" parentheses mean the endpoint is not included (open interval).
• For example, \( [a, b) \) includes all numbers from \( a \) to \( b \), including \( a \) but excluding \( b \).

Interval notation helps in simplifying the representation of solution sets, making them easier to read and understand. In trigonometric problems, it often helps in ensuring solutions meet specific range requirements, like ensuring angle measures fit within one full rotation of the unit circle.