91Ó°ÊÓ

To solve the equation \( \sin^2 x - \sin x - 1 = 0 \), we recognize it as a quadratic equation, but instead of a variable like \( x \) or \( y \), it is in terms of \( \sin x \). Quadratic equations have the general form \( ax^2 + bx + c = 0 \). In this exercise, replacing \( \sin x \) by \( y \) simplifies the process, transforming the equation to \( y^2 - y - 1 = 0 \). This step is crucial because it reduces the problem to a familiar type of equation called a quadratic equation, which is solvable by standard methods. Solving quadratic equations:

• Use the quadratic formula: \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
• Substitute values for \( a \), \( b \), and \( c \) to find solutions.
• Here, \( a = 1 \), \( b = -1 \), and \( c = -1 \), resulting in \: \( y_1 \approx 1.6180 \) and \( y_2 \approx -0.6180 \).

The goal is to find valid solutions where \( y = \sin x \) lies within its range, \([-1, 1]\). Understanding these basic methods equips you to tackle trigonometric equations by using known algebraic techniques.

The sine function is a periodic function with a range from -1 to 1. This characteristic plays a fundamental role in solving trigonometric equations like \( \sin^2 x - \sin x - 1 = 0 \). Upon finding that \( \sin x \approx -0.6180 \), we use the inverse sine function to solve for the angle \( x \). Inverse sine function:

• \( \arcsin(y) \) is utilized to find \( x \), the angle where the sine of the angle equals \( y \).
• It yields a principal value from \([-\frac{\pi}{2}, \frac{\pi}{2}]\), that may need adjustment to fit within the required range.

For \( \sin x = -0.6180 \), the basic solution found by \( \arcsin(-0.6180) \approx -0.6660 \) radians indicates an angle which needs to be adapted to fall within the specified interval \([0, 2\pi)\). By understanding these properties, one can find all possible solutions and adjust them accordingly.

After using the inverse sine function to find \( x \approx -0.6660 \), we need to find solutions in multiple angles where the sine function achieves the value \(-0.6180\). Identifying solutions:

• The angle -0.6660 is the result of the inverse calculation but falls outside the given interval \([0, 2\pi)\).
• Adjust it by adding \(2\pi\) to shift it within the required range, resulting in \( x_1 \approx 5.6171 \).
• Remember that sine is negative in both the third and fourth quadrants. Find solutions corresponding to these quadrants.
• For the third quadrant, compute \( \pi - (-0.6660) \approx 3.8072 \) as \( x_2 \).

This step-by-step approach ensures that all valid angle solutions are identified within the specified range, \([0, 2\pi)\). Understanding when and how to adjust angles is crucial in trigonometric problem solving.

Interval notation offers a concise way to share the range of possible solutions or numbers. In our exercise, solutions must fit within the interval \([0, 2\pi)\). The notation \([0, 2\pi)\) means:

• "[0," indicates inclusion of 0 in the solution set.
• "2\pi)" signifies exclusion of the endpoint \(2\pi\) itself.

Hence, solutions are found strictly between zero up to but not including \(2\pi\). Checking solutions, such as \(x_1 \approx 5.6171\) and \(x_2 \approx 3.8072\), against this interval guarantees they fit the problem's requirements. Mastering interval notation helps accurately communicate solution ranges and ensure calculations stay within problem constraints.