Problem 78 - Algebra and Trigonometry

Question

Use an identity to solve each equation on the interval \([0,2 \pi)\) $$ \sin x+\cos x=-1 $$

Step-by-Step Solution

Answer

The equation \(\sin x + \cos x = -1\) has no solutions on the interval \([0,2 \pi)\).

Step 1: Rewrite the equation

Rewrite the given equation \(\sin x + \cos x = -1\) using the Pythagorean identity to gain a quadratic equation. For this to be easier to manipulate, denote \(\cos x = p\). This gives \(\sin x = \sqrt{1-p^2}\) considering only the positive square root as the negative square root would yield no solution inside the given interval. Substitute these into the original equation to get \(\sqrt{1-p^2} + p = -1\) which simplifies to \(-p^2 - p - 2 = 0\).

Step 2: Solve the quadratic equation

Set up the quadratic equation in standard form and solve for \(p\), the equivalent of \(\cos x\). The quadratic equation \(-p^2 - p - 2 = 0\) can be rewritten to \(p^2 + p + 2 = 0\). Solve this using the quadratic formula \(\frac{-b \pm \sqrt{b^2-4ac}}{2a}\) yielding \(p = \frac{-1 \pm \sqrt{1-4*2}}{2}\). This does not have a real solution, indicating that the original equation \(\sin x + \cos x = -1\) has no solution in the interval \([0,2 \pi)\).

Step 3: Interpret the results

No real solutions for \(p\) mean that there are no solutions for \(\sin x + \cos x = -1\) in the interval \([0,2 \pi)\). Thus the given equation has no solutions in interval \([0,2 \pi)\).

Key Concepts

Pythagorean Identity

The Pythagorean identity is a fundamental relationship in trigonometry, which expresses a link between the sine and cosine of an angle. The identity is typically written as

\[\sin^2 x + \cos^2 x = 1\]

This equation is derived from the Pythagorean theorem relating the sides of a right-angled triangle. In the context of solving trigonometric equations, this identity is incredibly useful because it allows us to express one trigonometric function entirely in terms of another.

For example, if we have an equation involving both \(\sin x\) and \(\cos x\), we can solve for one of these variables and then use the Pythagorean identity to write everything in terms of the other variable. This can greatly simplify the equation and facilitate finding solutions.

Quadratic Equations

Quadratic equations, which take the general form

\[ax^2 + bx + c = 0\]

are polynomial equations of degree two. They are characterized by three coefficients: \(a\), \(b\), and \(c\), with \(a eq 0\). One of the most powerful tools for solving these equations is the quadratic formula:

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

The value under the square root, \(b^2 - 4ac\), is called the discriminant and it determines the nature of the solutions. If the discriminant is positive, the equation has two distinct real solutions; if it's zero, there's exactly one real solution; and if it's negative, as is the case in our problem, the equation has no real solutions. Understanding how to manipulate and solve quadratic equations is essential in algebra and appears frequently in trigonometry when dealing with trigonometric identities.

No Solution in Trigonometry

Situations where a trigonometric equation has no solution often occur and recognizing them is an important skill. One common reason for 'no solution' is when the equation involves the sum of a sine and cosine function equating to a value outside their possible range. Since both \(\sin x\) and \(\cos x\) have values between \(-1\) and \(1\), their sum cannot be less than \(-2\) or more than \(2\).

The exercise provided results in a quadratic equation where the discriminant is negative. This scenario indicates that there are no real values of \(p\), and by extension, no real angle \(x\) within the given interval that can satisfy the original trigonometric equation. Understanding that some equations naturally have no solution due to the inherent properties of trigonometric functions helps avoid confusion and saves time when solving these types of problems.