Problem 40 - Algebra and Trigonometry

Question

Involve trigonometric equations quadratic in form. Solve each equation on the interval \([0,2 \pi)\) $$ 2 \sin ^{2} x+\sin x-1=0 $$

Step-by-Step Solution

Answer

Substitute \( p = \sin x \) to turn this into a quadratic equation. Solve for \( p \) and back-substitute to find the different values of \( x \) in the interval \([0, 2\pi)\).

Step 1: Convert to Quadratic Equation Format

First, let's convert this trigonometric equation into a quadratic equation as it's easier to work with. Let \( p = \sin x \). Substituting this into the equation we get \( 2p^2 + p - 1 = 0 \)

Step 2: Solving the Quadratic Equation

Now apply the quadratic formula where \(a=2\), \(b=1\), and \(c=-1\). The solutions will be \( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), which simplifies to \( p = \frac{-1 \pm \sqrt{5}}{4} \). Thus the solutions are \( p_1 = \frac{-1 + \sqrt{5}}{4} \) and \( p_2 = \frac{-1 - \sqrt{5}}{4} \)

Step 3: Get the values of x

These are the solutions for \( p = \sin x \). Solve for x by finding the inverse sine for \( p_1 \) and \( p_2 \). The solutions in the interval \([0,2 \pi)\) will be \( x = \sin^{-1}(p_1), \pi - \sin^{-1}(p_1), \sin^{-1}(p_2), \pi - \sin^{-1}(p_2) \)

Step 4: Verifying the X values

Finally, verify that all the x values are indeed in the interval \([0,2 \pi)\). If any value is outside this interval, it will be discarded

Key Concepts

Quadratic Equations

Quadratic equations form the foundation of many mathematical problems, including some trigonometric equations. A quadratic equation is generally expressed in the form \( ax^2 + bx + c = 0 \). By substituting \( \sin x \) with a variable \( p \), the trigonometric equation \(2 \sin^2 x + \sin x - 1 = 0\) becomes \(2p^2 + p - 1 = 0\). Working with this algebraic form often simplifies the problem-solving process, allowing us to use tools like the quadratic formula. Remember, solving a quadratic equation involves finding the values of \( x \) that satisfy the equation, which can often have two solutions based on the quadratic formula.

Interval Notation

Interval notation is a simple way to express the range of solutions we're interested in, especially when solving equations within set bounds. For instance, the interval \([0, 2\pi)\) means all values starting at zero, up to but not including \(2\pi\). This concept is important in trigonometry, as we often want to find solutions for angles within one complete cycle of the unit circle. When solving equations, ensure that the solutions fall within the specified interval. If any solution exceeds this interval, it needs to be adjusted or discarded.

Inverse Trigonometric Functions

Inverse trigonometric functions are essential for finding angles from known trigonometric values. When you solve \( \sin x = p_1 \), you are actually looking for an angle \( x \) whose sine is \( p_1 \). Using \( \sin^{-1}(p) \), you can determine this angle. Remember that inverse trigonometric functions can yield multiple results because trigonometric functions are periodic. In this exercise, we explored both \( \sin^{-1}(p) \) and \( \pi - \sin^{-1}(p) \), as they both represent possible angles within our interval, highlighting the function's periodic nature.

Solving Equations

Solving equations involves a series of steps to find the unknown values that satisfy them. Starting with the given trigonometric equation, the exercise transformed it into a quadratic equation. After finding the solution \( p \), using the quadratic formula, substitute back to solve for original values like \( x \). This often involves performing additional calculations using inverse trigonometric functions and checking solution validity against given intervals. Always verify that your solution meets all given criteria before finalizing it. Careful attention to each step helps ensure accuracy and completeness.