Problem 62 - Precalculus: Functions and Graphs

Question

Use inverse trigonometric functions to find the solutions of the equation that are in the given interval, and approximate the solutions to four decimal places. \(6 \sin ^{2} x=\sin x+2\) \([0,2 \pi)\)

Step-by-Step Solution

Answer

Solutions are \(x \approx 0.7297, 5.5535, 0.5236, 2.6180\) in \([0, 2\pi)\).

Step 1: Rearrange the Equation

The given equation is \(6 \sin^2 x = \sin x + 2\). Let's rearrange it to become a standard quadratic equation in terms of \(\sin x\).Subtract \(\sin x + 2\) from both sides:\[6 \sin^2 x - \sin x - 2 = 0\]

Step 2: Factor the Quadratic

Now, let's factor the quadratic equation \(6 \sin^2 x - \sin x - 2 = 0\).The equation can be factored as:\((3 \sin x + 2)(2 \sin x - 1) = 0\)

Step 3: Solve the Factored Equations

From the factored equation \((3 \sin x + 2)(2 \sin x - 1) = 0\), we get two equations:1. \(3 \sin x + 2 = 0\)2. \(2 \sin x - 1 = 0\)Let's solve each one.

Step 4: Solve First Equation

For the first equation, \(3 \sin x + 2 = 0\):\[3 \sin x = -2\]\[\sin x = -\frac{2}{3}\]This value is within the range of the sine function, so we can find solutions for \(x\) using the inverse sine function within the given interval \([0, 2\pi)\).

Step 5: Solve Second Equation

For the second equation, \(2 \sin x - 1 = 0\):\[2 \sin x = 1\]\[\sin x = \frac{1}{2}\]Use the inverse sine function to solve for \(x\) within the interval \([0, 2\pi)\).

Step 6: Find Solutions from First Equation

For \(\sin x = -\frac{2}{3}\), the solutions within \([0, 2\pi)\) correspond to angles where the sine is negative. We find these solutions using:\[x = \arcsin\left(-\frac{2}{3}\right)\] and \[x = \pi - \arcsin\left(-\frac{2}{3}\right)\]Calculate these angles using a calculator to approximate to four decimal places:\[x_1 \approx 0.7297\]\[x_2 \approx 5.5535\] (This is obtained using \(2\pi - x_1\))

Step 7: Find Solutions from Second Equation

For \(\sin x = \frac{1}{2}\), the solutions within \([0, 2\pi)\) are:1. \(x = \frac{\pi}{6}\)2. \(x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}\)Approximate these values:\[x_3 \approx 0.5236\]\[x_4 \approx 2.6180\]

Step 8: Confirming the Interval

Ensure that all the found solutions \(x_1 \approx 0.7297\), \(x_2 \approx 5.5535\), \(x_3 \approx 0.5236\), and \(x_4 \approx 2.6180\) are indeed within the interval \([0, 2\pi)\), which they are.

Key Concepts

Quadratic Equation in Trigonometry

In trigonometry, a quadratic equation may involve trigonometric functions. In this particular exercise, we see an equation involving the sine function: \(6 \sin^2 x = \sin x + 2\). Our first task is to rearrange this into a standard quadratic form.
To achieve this, we subtract \(\sin x + 2\) from both sides, leading to the equation:

• \(6 \sin^2 x - \sin x - 2 = 0\)

Now, it resembles a typical quadratic equation where \(\sin x\) is our variable. This transformation allows us to apply techniques we already know, like factoring or using the quadratic formula.
The goal is to express the trigonometric equation in terms of a single trigonometric function, which makes it easier to solve using familiar algebraic techniques.

Solving Trigonometric Equations

After transforming our trigonometric equation into a quadratic form, we proceed to solve it. Solving trigonometric equations can sometimes be tricky, but factoring or using the quadratic formula offers a systematic path.
In this exercise, the equation \(6 \sin^2 x - \sin x - 2 = 0\) can be factored into two separate expressions:

• \((3 \sin x + 2)(2 \sin x - 1) = 0\)

From here, we split the problem into two simpler equations:
1. \(3 \sin x + 2 = 0\)
2. \(2 \sin x - 1 = 0\)
We solve each for \(\sin x\):

• For \(3 \sin x + 2 = 0\), rearranging gives \(\sin x = -\frac{2}{3}\).
• For \(2 \sin x - 1 = 0\), rearranging gives \(\sin x = \frac{1}{2}\).

Once you find the values of \(\sin x\), you can use inverse trigonometric functions to determine the actual angle solutions within the specified interval \([0, 2\pi)\).

Sine Function

The sine function is one of the basic trigonometric functions. It relates the angle in a right triangle to the ratio of the length of the opposite side over the hypotenuse.
When solving for \(x\) in equations involving the sine function, we often need to use the inverse sine function. This gives us angles corresponding to a particular sine value.
For \(\sin x = -\frac{2}{3}\), we use \(x = \arcsin(-\frac{2}{3})\) to find the angle in the correct quadrant, aware that the sine function is negative in the third and fourth quadrants.

• The solutions will be approximated as \(0.7297\) and \(5.5535\) based on the interval \([0, 2\pi)\).

For \(\sin x = \frac{1}{2}\), the reference angles are well-known:

• \(x_1 = \frac{\pi}{6}\) or approximately \(0.5236\)
• \(x_2 = \frac{5\pi}{6}\) or approximately \(2.6180\)

In each case, solutions are carefully identified to fit within the given interval of the problem, always checking both positive or negative values based on the function's periodicity and symmetry.