Problem 69 - Algebra and Trigonometry

Question

Solve: \(\cos ^{2} x+\sin x+1=0,0 \leq x \leq 2 \pi\) (Section \(6.5, \text { Example } 7)\)

Step-by-Step Solution

Answer

The equation \(\cos ^{2} x+\sin x+1=0\) has no solutions for \(x\) in the range \(0 \leq x \leq 2 \pi\).

Step 1: Use the Trigonometric Identity

To simplify the equation, the Pythagorean identity \(\cos^2x + \sin^2x = 1\) can be used. Since \(\cos^2x = 1 - \sin^2x\), the original equation can be rewritten as: \(1 - \sin^2x + \sin x +1 = 0\).

Step 2: Rewrite the Equation

Rewriting the equation gives \(\sin^2x - \sin x +2 = 0\).

Step 3: Solve for sin(x) as Quadratic Equation

Now, solve the quadratic equation \(\sin^2x - \sin x +2 = 0\) as if \(\sin x\) is a variable. As it turns out, this equation does not have real roots, thus there are no solutions for \(\sin x\).

Step 4: Check for All Possible Solutions

Since equation does not have real roots, there are no solutions for x in the range \(0 \leq x \leq 2 \pi\).

Key Concepts

Understanding the Pythagorean Identity

The Pythagorean identity is one of the cornerstones of trigonometry and plays a pivotal role in simplifying and solving trigonometric equations. It is based on the Pythagorean theorem, which states that in a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides. In the context of trigonometry, the Pythagorean identity is \[\begin{equation}\textstyle \begin{aligned} \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \sin^2x + \cos^2x = 1\textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \textstyle \end{aligned}\end{equation}\]This trigonometric formula expresses a fundamental relationship between the sine and cosine functions. It's handy when you encounter terms involving both \(\sin x\) and \(\cos x\) in an equation. You can often use the Pythagorean identity to replace one function with the other, thus simplifying the equation and making it easier to solve, as demonstrated in the current example.

When we encounter the equation \(\cos^2 x + \sin x + 1 = 0\), using the Pythagorean identity allows us to substitute for \(\cos^2 x\) and rewrite the equation as a function of \(\sin x\) alone. This substitution paves the way for applying techniques from algebra, such as solving quadratic equations, to trigonometric expressions.

Tackling Quadratic Equations in Trigonometry

Quadratic equations play a significant role in mathematics, extending their utility to the field of trigonometry. A quadratic equation is typically written as \(ax^2 + bx + c = 0\) and features a variable raised to the second power. In trigonometric problems, the variable is often a trigonometric function like \(\sin(x)\) or \(\cos(x)\).

Solving a trigonometric expression cast as a quadratic equation starts with identifying the trigonometric function as the variable. The standard quadratic formula—\(\frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\)—can then be applied, but with slight caution: since we're dealing with trigonometric functions, the solution has to be within the function's range. However, it's important to note that not all quadratic equations will have real solutions. If the discriminant (\(b^2 - 4ac\)) is negative, the equation has no real solutions, which is what we find in the original exercise \(\sin^2x - \sin x + 2 = 0\).

In the context of trigonometric equations, having no real solutions implies that there are no angles \(x\) which satisfy the given equation within the defined domain, usually between \(0\) and \(2\pi\) when dealing with radians, or \(0\) and \(360\degree\) for degrees.

Solving Trigonometric Equations Process

The process of solving trigonometric equations can range from straightforward to complex, depending on the form of the equation. The general approach follows these steps:

• Isolation: Aim to isolate the trigonometric function, if possible. Reduce the equation to a basic form where the function appears on one side, and a numerical value appears on the other.
• Use identities: Apply trigonometric identities, like the Pythagorean identity, to express the equation in terms of one function. This is crucial for simplifying and making the equation more manageable.
• Apply algebra: If the expression forms a quadratic equation, you can utilize algebraic methods such as factoring, or the quadratic formula to find the solutions for the trigonometric function.
• Check the domain: Remember to consider the domain of the function. For real functions, the solutions must be real numbers within the permissible range of the function.
• Analyze the solutions: Verify the solutions by plugging them back into the original equation to ensure their validity in the given range. Sometimes, additional solutions may arise from the periodic nature of trigonometric functions.

It's vital to understand that even if you arrive at a valid trigonometric function value after solving the equation, there may be no corresponding angle that satisfies the particular constraints of the problem. This is the case in the given exercise, where after simplifying and attempting to solve the quadratic equation, it's discovered that there are no real solutions for \(\sin x\) in the range of \(0\) to \(2\pi\). This result teaches an important lesson: not every trigonometric equation has a solution, and it's essential to identify such situations correctly.