Problem 96 - Precalculus Essentials

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 \(5 \sin ^{2} x-1=0\) on the interval [0,2\(\pi\))] are \(x \approx 0.4909, 2.6507, 3.6320, 5.7920\)

Step 1: Write equation in standard form

Rewrite the given equation in standard form: \(5 \sin ^{2} x = 1\), which simplifies to \(\sin^2 x = 0.2\)

Step 2: Solving for sin x

Find the value of \(\sin x\) by taking the square root on both sides: \(\sin x= \sqrt{0.2} \) or \(\sin x = -\sqrt{0.2}\).

Step 3: Finding values for x

Solve the equation \(\sin x= \sqrt{0.2} \) using a calculator. Make sure the calculator is in radian mode. This give us two solutions \(x\approx 0.4909\) and \(x\approx 2.6507\) in the interval \([0,2 \pi)\). Similarly, for \(\sin x = -\sqrt{0.2}\), the solutions are \(x\approx 3.6320\) and \(x\approx 5.7920\) in the interval \([0, 2\pi)\).

Step 4: Checking the solutions

Substitute the solutions back into the original equation to check if they satify the conditions. After substituting, it can be verified that all these solutions are correct.

Key Concepts

Sine function

The sine function is fundamental in trigonometry and defines how we measure angles and their relationships in terms of a unit circle. Think of it as associating each angle with a point on the circle. In simple terms, for a given angle \( x \), the sine function, \( \sin x \), gives the y-coordinate of that point on the unit circle.

Key points about the sine function include:

• It is periodic with a period of \( 2\pi \), meaning it repeats every \( 2\pi \) radians.
• It ranges from -1 to 1, as the maximum and minimum values of \( \sin x \) are 1 and -1, respectively.
• It is a smooth and continuous function, making it very useful for modeling periodic phenomena such as sound and light waves.

In the context of solving the given equation, \( \sin^2 x = 0.2 \), we are interested in finding the angle \( x \) that satisfies this equation. It means finding where the sine of our angle squared equals 0.2, which directly informs us about possible solutions within the given interval \([0, 2\pi)\).

Radians

Radians are a crucial unit of angle measurement used in trigonometry. Unlike degrees, which divide a circle into 360 parts, radians consider the circle's arc. One radian is the angle covered when the length of the arc is equal to the radius of the circle.

Some essential points about radians:

• The total angle in a circle is \( 2\pi \) radians, equivalent to 360 degrees.
• This makes \( \pi \) radians equal to 180 degrees, helping you convert between radians and degrees.
• Radians are more natural for mathematics, as they simplify derivative and integral calculations, especially in calculus.

When we solve trigonometric equations like \( 5 \sin^2 x - 1 = 0 \) over the interval \([0, 2\pi)\), we specifically look for solutions in radian measure. Calculators need to be in radian mode to provide accurate solutions for angles, which is why our calculated solutions conform to the radians unit.

Square root property

The square root property is a useful algebraic tool when solving equations that involve squares. When you have an equation of the form \( a^2 = b \), you solve it by taking the square root of both sides, leading to \( a = \sqrt{b} \) or \( a = -\sqrt{b} \). This property allows us to find all possible values of the variable.

In this exercise, we apply the square root property to \( \sin^2 x = 0.2 \). It effectively simplifies our work:

• By taking the square root of both sides, we get two potential solutions: \( \sin x = \sqrt{0.2} \) and \( \sin x = -\sqrt{0.2} \).
• Since sine can be positive or negative due to its range, we must consider both solutions for completeness.
• Solving these gives specific x-values within our interval, ensuring all potential angles are discovered.

The square root property here ensures we don't miss any solutions, as both the positive and negative roots lead to valid sine values, rounding out our solution set.