Problem 86 - Algebra and Trigonometry Real Mathematics, Real People

Question

Use inverse functions where necessary to solve the equation. $$2 \cos ^{2} x+7 \sin x=5$$

Step-by-Step Solution

Answer

Solving the equation involves use of trigonometric identities to express the equation in terms of one trigonometric function (\( \sin x \) in this case), solving the resultant quadratic equation to get the possible values of \( \sin x \), and finally using inverse trigonometric function to solve for \( x \). The exact solution will depend on the roots of the quadratic equation.

Step 1: Create an Expression Containing Only One Trigonometric Function

To simplify the calculation, we can use the Pythagorean identity of trigonometric functions, \(\cos ^{2} x = 1 - \sin ^{2} x\). Substitute this into the given equation to express everything in terms of \(\sin x\).

Step 2: Formulate a Quadratic Equation

Substituting \(\cos ^{2} x = 1 - \sin ^{2} x\) into the original equation \(2 \cos ^{2} x+7 \sin x=5\) we get \(2(1 - \sin ^{2} x) + 7 \sin x - 5 = 0\). Simplifying this gives \(2 \sin ^{2} x - 7 \sin x + 3 = 0\), which can be considered as a quadratic equation with \( \sin x\) as variable. So the task essentially becomes solving the quadratic equation.

Step 3: Solving the Quadratic Equation

Using the quadratic formula \(\frac{-b \pm \sqrt{b^2-4ac}}{2a}\) we just get the solutions for \(\sin x\), which we will denote as \(s_1\) and \(s_2\). These will be the roots of the equation \(2 \sin ^{2} x - 7 \sin x + 3 = 0\).

Step 4: Using Inverse Trigonometric Function to Solve for x

Once we have the roots, we can use the arcsine function (inverse of sine function) to solve for \(x\). We need to consider the domain of the arcsine function while doing so, and two sets of solutions can be obtained depending on whether \(x\) falls in \([-\pi/2, \pi/2]\) or \([\pi/2, 3\pi/2]\).

Step 5: Final Solution

After following all the above steps, the solution values for \(x\) can be obtained. It should be noted that the solution might not be unique due to the periodic nature of trigonometric functions, as multiple angles can produce the same sine value.

Key Concepts

Pythagorean Identity

The Pythagorean identity is a fundamental relation in trigonometry. It states that for any angle \( x \): \[ \cos^2 x + \sin^2 x = 1 \] This identity connects the sine and cosine functions. It can be used to express one function in terms of the other.

• This can simplify equations involving trigonometric functions.
• In our problem, we express \( \cos^2 x \) in terms of \( \sin^2 x \) using \( \cos^2 x = 1 - \sin^2 x \).
• This substitution helps us write the equation in terms of a single trigonometric function, \( \sin x \).

Understanding and using the Pythagorean identity is crucial when simplifying trigonometric equations.

Inverse Trigonometric Functions

Inverse trigonometric functions allow us to find the angles corresponding to specific trigonometric ratios. They are crucial when solving trigonometric equations. The commonly used inverse functions are:

• \( \sin^{-1} \)
• \( \cos^{-1} \)
• \( \tan^{-1} \)

These functions are confined to specific ranges:

• The range of \( \sin^{-1} \) (arcsine) is \([-rac{\pi}{2}, \frac{\pi}{2}]\).
• This range limitation ensures that each inverse trigonometric function is a true function (no two different angles have the same ratio in its range).

In problems like ours, once \( \sin x \) values are found, the inverse sine function helps retrieve the angle \( x \).

Arcsine Function

The arcsine function \( \sin^{-1} \) is the inverse of the sine function. Given a sine value, \( \sin^{-1} \) returns the corresponding angle.

• The function maps from \([-1, 1]\) (sine values) to \([-rac{\pi}{2}, \frac{\pi}{2}]\) (angles).
• This restriction ensures the function is one-to-one, providing a unique angle for each sine value.

When solving trigonometric equations using arcsine, remember:

• Check if the sine value is in the valid range \([-1, 1]\).
• Use the principal value returned by the arcsine function first, then consider the periodic properties of sine.

In our specific problem, after finding \( \sin x \) values, we use \( \sin^{-1} \) to solve for \( x \). Always consider other angles for possible solutions due to the periodic nature of the sine function.

Solving Quadratic Equations

Solving quadratic equations is a fundamental algebraic process. A quadratic equation has the form:\[ ax^2 + bx + c = 0 \]To solve it, we can use:

• Factoring, if the equation can be easily decomposed into simple binomials.
• The quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), which provides solutions for any solvable quadratic equation.

In our exercise, after substituting \( \cos^2 x = 1 - \sin^2 x \) and simplifying, we derived a quadratic equation involving \( \sin x \):\[ 2\sin^2 x - 7\sin x + 3 = 0 \]Using the quadratic formula:

• First, we determine the discriminant \( b^2 - 4ac \) to ensure real solutions exist.
• We then calculate the possible \( \sin x \) values using the quadratic formula.
• These solutions give us \( \sin x \) which can then be used in the arcsine function to find \( x \).

Always remember to check both positive and negative roots, ensuring all possible solutions for the original trigonometric equation are considered.