Problem 96 - Algebra and Trigonometry

Question

Use a calculator to solve each equation, correct to four decimal places, on the interval \([0,2 \pi)\) $$ 5 \sin ^{2} x-1=0 $$

Step-by-Step Solution

Answer

The solutions to the equation on the interval \([0,2 \pi)\) are \(x_1 = \arcsin(\sqrt{1/5})\) and \(x_2 = \pi - \arcsin(\sqrt{1/5})\) for \(\sin x =\sqrt{\frac{1}{5}}\), and \(x_3 = \arcsin(-\sqrt{1/5}) + \pi\) and \(x_4 = 2 \pi - \arcsin(\sqrt{1/5})\) for \(\sin x =-\sqrt{\frac{1}{5}}\). Remember to use a calculator to get the values correct to four decimal places.

Step 1: Simplify the Equation

The given equation is \(5 \sin ^{2} x - 1 = 0\). First, isolate the \(\sin ^{2} x\). To do this, add 1 to both sides and then divide by 5. This yields: \(\sin ^{2} x = \frac{1}{5}\).

Step 2: Apply the Square Root

Since \(\sin ^{2} x =\frac{1}{5}\), then \(\sin x =\sqrt{\frac{1}{5}}\) or \(\sin x =-\sqrt{\frac{1}{5}}\). So, we have two possible values.

Step 3: Solve for \(x\)

Apply the inverse sine function to both sides of the equations \[\sin x =\sqrt{\frac{1}{5}}\] and \[\sin x =-\sqrt{\frac{1}{5}}\] to find the values for \(x\). Make sure the values fall within the interval \([0,2 \pi)\). Use the calculator to get the accurate values.

Step 4: Verify the Solution

Finally, substitute the values obtained for \(x\) back into the original equation to verify that they are indeed solutions.

Key Concepts

Understanding the Inverse Sine Function

Trigonometric equations, such as the given example with sine, often require the use of inverse functions to solve for the angle. The inverse sine function, also known as arcsine, is written as \( \sin^{-1} x \) or \( \text{asin}(x) \). This function helps us find an angle whose sine is a given number.

For any value \( y \) within the range \(-1\) to \(1\), the inverse sine function returns an angle within \([-\frac{\pi}{2}, \frac{\pi}{2}]\). However, we are tasked with finding solutions in the interval \([0, 2\pi)\), so it’s essential to consider other possible angles where the sine of an angle equals \( y \).

When solving equations, remember that sine is positive in the first and second quadrants, and negative in the third and fourth quadrants. This characteristic of sine helps determine additional solutions within your specified interval.

Interval Notation Explained

Interval notation is a mathematical notation used to represent a range of values, specified by a pair of numbers. In the exercise, we work within the interval \([0, 2\pi)\).

Intervals can be open or closed. A closed interval includes its endpoints, denoted with square brackets like \([a, b]\), which means \(a \leq x \leq b\). An open interval, denoted with parentheses, does not include its endpoints, like \((a, b)\), meaning \(a < x < b\).

Our problem uses a closed bracket at \(0\) and an open bracket at \(2\pi\), thus including \(0\) but excluding \(2\pi\). It’s crucial because sine functions are periodic, and within one full cycle, \(2\pi\) is not included to avoid duplication of the starting point, \(0\).

The Process of Solving Equations

When solving trigonometric equations, the goal is to find the angle(s), or \(x\), within a given range that satisfy the equation. Our equation, after simplification, becomes \( \sin^2 x = \frac{1}{5} \).

This equation implies two possibilities for sine: \( \sin x = \sqrt{\frac{1}{5}} \) and \( \sin x = -\sqrt{\frac{1}{5}} \). Each case needs to be solved separately using the inverse sine function.

• Use \( \sin^{-1}(\sqrt{\frac{1}{5}}) \) to find an initial angle. Calculate additional angles in relevant quadrants where sine is positive.
• For \( \sin x = -\sqrt{\frac{1}{5}} \), use \( \sin^{-1}(-\sqrt{\frac{1}{5}}) \) and find corresponding angles in quadrants where sine is negative.

Utilize the calculator to determine these angles precisely, ensuring they align within the defined interval \([0, 2\pi)\).

Importance of Verifying Solutions

After calculating possible angles, verification of solutions is a critical step in solving equations. This checks the correctness of the mathematical process and the accuracy of the technological tools, like calculators, involved.

Verification involves substituting each calculated angle back into the original equation. You confirm these values solve the equation by ensuring both sides equal after substitution.

• If \( x = 3D \) satisfies \( 5 \sin^2(x) - 1 = 0 \), the equation simplifies correctly.
• If a value does not satisfy the equation, it may fall out of the specified interval or is a potential calculation error.

Verification helps ensure complete understanding and accuracy in solving trigonometric equations, providing confidence in your solutions every step of the way.