Key Concepts and Formulas

• Reciprocal Trigonometric Identity: $\csc \theta = \frac{1}{\sin \theta}$
• Value of $\sin 18^\circ$: $\sin 18^\circ = \frac{\sqrt{5}-1}{4}$
• Rationalization: Multiplying the numerator and denominator of a fraction by the conjugate of the denominator to eliminate radicals in the denominator.

Step-by-Step Solution

Step 1: Find $\csc 18^\circ$ using the reciprocal identity.

We are given $\sin 18^\circ = \frac{\sqrt{5}-1}{4}$ and we need to find $\csc 18^\circ$. We use the reciprocal identity $\csc \theta = \frac{1}{\sin \theta}$.

$\csc 18^\circ = \frac{1}{\sin 18^\circ} = \frac{1}{\frac{\sqrt{5}-1}{4}} = \frac{4}{\sqrt{5}-1}$

Explanation: We substitute the value of $\sin 18^\circ$ into the reciprocal identity to express $\csc 18^\circ$.

Step 2: Rationalize the denominator of $\csc 18^\circ$.

To simplify the expression, we rationalize the denominator by multiplying the numerator and denominator by the conjugate of $\sqrt{5}-1$, which is $\sqrt{5}+1$.

$\csc 18^\circ = \frac{4}{\sqrt{5}-1} \cdot \frac{\sqrt{5}+1}{\sqrt{5}+1} = \frac{4(\sqrt{5}+1)}{(\sqrt{5})^2 - 1^2} = \frac{4(\sqrt{5}+1)}{5-1} = \frac{4(\sqrt{5}+1)}{4} = \sqrt{5}+1$

Explanation: Rationalizing the denominator eliminates the radical from the denominator, making the expression simpler. We use the difference of squares factorization: $(a-b)(a+b) = a^2 - b^2$.

Step 3: Set $x = \csc 18^\circ$.

Let $x = \csc 18^\circ$. Then, we have:

$x = \sqrt{5} + 1$

Explanation: We substitute $x$ for $\csc 18^\circ$ to form an equation in terms of $x$.

Step 4: Isolate the radical term.

To eliminate the radical, we isolate the $\sqrt{5}$ term.

$x - 1 = \sqrt{5}$

Explanation: Isolating the radical before squaring is a crucial step. If we squared the original equation directly, we'd still have a radical term.

Step 5: Square both sides of the equation.

Squaring both sides of the equation $x-1 = \sqrt{5}$, we get:

$(x-1)^2 = (\sqrt{5})^2$ $x^2 - 2x + 1 = 5$

Explanation: Squaring both sides eliminates the square root. We expand $(x-1)^2$ correctly using the binomial formula.

Step 6: Rearrange the equation into the standard quadratic form.

Rearranging the terms to get the standard quadratic form $ax^2 + bx + c = 0$, we have:

$x^2 - 2x + 1 - 5 = 0$ $x^2 - 2x - 4 = 0$

Explanation: We rearrange the terms to obtain a standard quadratic equation where $\csc 18^\circ$ is a root.

Common Mistakes & Tips

• Memorize $\sin 18^\circ$: Remembering the value of $\sin 18^\circ$ is extremely helpful for these types of problems.
• Correctly Square Binomials: Double-check your expansion of $(a \pm b)^2$. The middle term is often missed.
• Isolate Radicals: Isolate the radical term before squaring to avoid having a radical in the resulting equation.

Summary

We used the reciprocal trigonometric identity and the known value of $\sin 18^\circ$ to find $\csc 18^\circ$. Then, we rationalized the denominator, set $x = \csc 18^\circ$, isolated the radical, and squared both sides to obtain a quadratic equation. The final quadratic equation for which $\csc 18^\circ$ is a root is $x^2 - 2x - 4 = 0$.

Final Answer

The final answer is \boxed{x^2 - 2x - 4 = 0}, which corresponds to option (D).