Problem 10 - Algebra and Trigonometry Real Mathematics, Real People

Question

Value is a solution of the equation. \(\csc ^{4} x-4 \csc ^{2} x=0\) (a) \(x=\frac{\pi}{6}\) (b) \(x=\frac{5 \pi}{6}\)

Step-by-Step Solution

Answer

The solutions to the equation \(\csc ^{4} x-4 \csc ^{2} x=0\) are \(x={\pi}/{6}, {5\pi}/{6}\).

Step 1: Write the Trigonometric Equation into a Polynomial Form

The given trigonometric equation is \(\csc ^{4} x-4 \csc ^{2} x=0\), which can be rewritten into a polynomial form as \((\csc^2 x)^2 - 4(\csc^2 x) = 0\). This is essentially a quadratic equation.

Step 2: Factorize the Quadratic Equation

We can treat \(\csc^2 x\) as one variable to simplify the equation. Let \(t = \csc^2 x\), so the equation becomes \(t^2 - 4t = 0\). Factor this equation gives \(t(t - 4) = 0\). From this, we get two solutions: \(t = 0\) and \(t = 4\). Here \(t\) represents \(\csc^2 x\). Therefore, the solutions of the original equation are \(\csc^2 x = 0\) and \(\csc^2 x = 4\).

Step 3: Solve for \(x\)

For the equation \(\csc^2 x = 0\), since \(\csc x = 1/ \sin x\), it is undefined at \(x = k\pi\), where \(k\) is an integer. Hence, there are no solutions to this equation.\n\nThen solve the equation \(\csc^2 x = 4 .\),1. We solve for \(\csc x\). So, \(\csc x = ±2\). 2. The equation becomes \(\sin x = ±1/2\). 3. This has solutions \(x={\pi}/{6}, {5\pi}/{6}\) for \(0 ≤ x < 2\pi\).

Step 4: Check the Solutions

The solutions must be checked to ensure that it satisfies the original equation. Therefore, the answer is \(x={\pi}/{6}, {5\pi}/{6}\).

Key Concepts

Polynomial Form

To solve trigonometric equations, it often helps to transform them into polynomial form. This simplifies solving by turning a complex trigonometric expression into a more familiar polynomial expression. Given the equation \(\csc^4 x - 4 \csc^2 x = 0\), you can identify it as equivalent to a quadratic polynomial by making a substitution. Let \(t = \csc^2 x\), then the equation becomes \(t^2 - 4t = 0\). This step makes recognizing and solving the pattern much more manageable, leading to techniques you might already be comfortable with from algebra, such as factorization.

Cosecant Function

The cosecant function, represented as \(\csc x\), is the reciprocal of the sine function: \(\csc x = \frac{1}{\sin x}\). This function tends to be less intuitive than sine or cosine because its values are undefined where the sine is zero. Understanding \(\csc x\) is crucial for solving the given equation since it involves powers of \(\csc x\). When converting trigonometric equations that involve cosecant into polynomial form, knowing this reciprocal relation is critical. Since \(\csc x\) is undefined when \(\sin x = 0\), be wary of checking these solutions, as no valid solution can exist where \(\csc x\) becomes undefined. Explore \(\csc x\)'s behavior and its graph for a better grasp of where the function can exist.

Quadratic Equation

Once an equation is in polynomial form, recognizing it as a quadratic equation is a tremendous advantage. Quadratic equations take the form \(ax^2 + bx + c = 0\) and are solvable using various methods, such as factorizing or using the quadratic formula. The equation \(t^2 - 4t = 0\) fits this format, making the next step straightforward. Solving this allows us to express the relationship between the variable \(t\) (which stands in for \(\csc^2 x\)) and the solutions. The nature of the quadratic (factorable in this instance) provides a gateway to quickly determine the roots and thus further translate these back into trigonometric solutions.

Factorization

Factorization is a method used to simplify polynomial equations by expressing them as products of simpler polynomials. Given the equation \(t^2 - 4t = 0\), it can be factorized as \(t(t - 4) = 0\). This gives us two distinct solutions: \(t = 0\) and \(t = 4\). Each factor represents a potential solution to the equation and translates back into the trigonometric space where \(t = \csc^2 x\). Note, however, that \(\csc^2 x = 0\) is not valid for our purposes since \(\csc x = \frac{1}{\sin x}\) is undefined if sine itself is zero. Thus, the practically viable solution remains \(\csc^2 x = 4\), guiding us to solve for \(x\).

Checking Solutions

Checking solutions is essential to verify the correctness of potential candidates derived in solving equations. For \(\csc x = \pm 2\), converting to sine we find \(\sin x = \pm \frac{1}{2}\). Solving this helps us identify specific solutions in the range \(0 \leq x < 2\pi\). Known solutions for \(\sin x = \frac{1}{2}\) include \(x = \frac{\pi}{6}\) and \(x = \frac{5\pi}{6}\). Double-check each solution against the original trigonometric equation to ensure their validity and accuracy. It's an important last step to avoid introducing extraneous solutions that might logically follow from algebra but don't actually satisfy the initial conditions of the trigonometric realm.