Solving $2 \sin ^2 Z+3 \sin Z+1=0$: A Step-by-Step Guide

When faced with trigonometric equations, especially those resembling quadratic forms, it's natural to feel a slight pause. However, with a systematic approach, these problems become quite manageable. The equation we're diving into today, $2 \sin ^2 z+3 \sin z+1=0$, is a perfect example of a quadratic trigonometric equation. This means it has a structure similar to a standard quadratic equation like $ax^2 + bx + c = 0$, but instead of a simple variable 'x', we have a trigonometric function, in this case, $\sin z$. Our goal is to find the values of $z$ that satisfy this equation. We'll break this down into logical steps, making it easy to follow along and understand the underlying principles. The first crucial step in tackling $2 \sin ^2 z+3 \sin z+1=0$ is to recognize its quadratic nature. By making a simple substitution, we can transform it into a form that is immediately familiar. Let's substitute $\sin z$ with a new variable, say $y$. This transforms our original equation into $2y^2 + 3y + 1 = 0$. This is a classic quadratic equation, and we have several methods at our disposal to solve it. The most common methods include factoring, completing the square, or using the quadratic formula. For $2 \sin ^2 z+3 \sin z+1=0$, factoring often presents a straightforward path if the expression can be factored neatly.

Let's explore the factoring method for our transformed equation, $2y^2 + 3y + 1 = 0$. We are looking for two numbers that multiply to $(2 \times 1) = 2$ and add up to $3$. The numbers $2$ and $1$ fit this criterion perfectly ($2 \times 1 = 2$ and $2 + 1 = 3$). Now, we can rewrite the middle term, $3y$, as $2y + y$. This allows us to factor by grouping:

$2y^2 + 2y + y + 1 = 0$

Group the terms:

$(2y^2 + 2y) + (y + 1) = 0$

Factor out the common terms from each group:

$2y(y + 1) + 1(y + 1) = 0$

Now, we can see a common binomial factor $(y + 1)$. Factor this out:

$(2y + 1)(y + 1) = 0$

For this product to be zero, at least one of the factors must be zero. This gives us two possible cases:

Case 1: $2y + 1 = 0$

Case 2: $y + 1 = 0$

Solving Case 1 for $y$: $2y = -1 \implies y = -1/2$

Solving Case 2 for $y$: $y = -1$

Now that we have found the possible values for $y$, we must substitute back $\sin z$ for $y$ to return to our original trigonometric equation, $2 \sin ^2 z+3 \sin z+1=0$. This step is crucial for finding the actual solutions for $z$. We have two sets of equations to solve:

1. $\sin z = -1/2$
2. $\sin z = -1$

These are now standard trigonometric equations that we can solve for $z$ within a typical range, usually $0 \le z < 2\pi$ (or 0 to 360 degrees), and then generalize the solutions.

Let's tackle $\sin z = -1/2$ first. We need to find angles $z$ whose sine is $-1/2$. We know that the sine function is negative in the third and fourth quadrants. The reference angle for which $\sin(\text{reference angle}) = 1/2$ is $\pi/6$ (or 30 degrees). Therefore, the solutions in the interval $0 \le z < 2\pi$ are:

In the third quadrant: $z = \pi + \pi/6 = 7\pi/6$

In the fourth quadrant: $z = 2\pi - \pi/6 = 11\pi/6$

So, for $\sin z = -1/2$, our solutions are $z = 7\pi/6$ and $z = 11\pi/6$ within one period.

Next, let's solve $\sin z = -1$. The sine function equals $-1$ at only one point within the interval $0 \le z < 2\pi$. This occurs at $z = 3\pi/2$ (or 270 degrees). This is a special case where the sine function reaches its minimum value.

So, within the primary interval $0 \le z < 2\pi$, the solutions to $2 \sin ^2 z+3 \sin z+1=0$ are $z = 7\pi/6$, $z = 11\pi/6$, and $z = 3\pi/2$. It's important to remember that trigonometric functions are periodic. This means that there are infinitely many solutions. To express the general solution, we add integer multiples of $2\pi$ to each particular solution, as the sine function has a period of $2\pi$.

The general solution for $\sin z = -1/2$ is:

$z = 7\pi/6 + 2n\pi$

$z = 11\pi/6 + 2n\pi$

where $n$ is any integer ($n \in \mathbb{Z}$).

The general solution for $\sin z = -1$ is:

$z = 3\pi/2 + 2n\pi$

where $n$ is any integer ($n \in \mathbb{Z}$).

Combining these, the complete set of solutions for the equation $2 \sin ^2 z+3 \sin z+1=0$ is represented by these general forms. It's always a good practice to check these solutions by substituting them back into the original equation to ensure accuracy. For example, let's check $z = 3\pi/2$: $\sin(3\pi/2) = -1$. Substituting into the equation: $2(-1)^2 + 3(-1) + 1 = 2(1) - 3 + 1 = 2 - 3 + 1 = 0$. This confirms $z = 3\pi/2$ is a valid solution. Similarly, checking $z = 7\pi/6$: $\sin(7\pi/6) = -1/2$. Substituting: $2(-1/2)^2 + 3(-1/2) + 1 = 2(1/4) - 3/2 + 1 = 1/2 - 3/2 + 1 = -2/2 + 1 = -1 + 1 = 0$. This also confirms the solution. The process of solving quadratic trigonometric equations like $2 \sin ^2 z+3 \sin z+1=0$ hinges on recognizing the quadratic form, using substitution to simplify it, solving the resulting algebraic equation, and then solving the basic trigonometric equations. Each step builds upon the last, leading you systematically to the complete set of solutions. Understanding the unit circle and the behavior of trigonometric functions in different quadrants is also fundamental to accurately finding the specific angle solutions. This methodical approach ensures that no solutions are missed and that the general solution correctly captures the periodic nature of the trigonometric functions. The ability to manipulate and solve equations of this nature is a cornerstone of advanced trigonometry and calculus, finding applications in various fields of science and engineering. It is a testament to the power of algebraic techniques applied to the cyclical world of trigonometry. The initial challenge of a seemingly complex equation like $2 \sin ^2 z+3 \sin z+1=0$ dissolves with a clear understanding of these fundamental steps, empowering you to tackle even more intricate problems.

For further exploration into trigonometric identities and equations, you can refer to resources like Khan Academy's Trigonometry Section, which offers comprehensive explanations and practice problems. Another excellent source for in-depth mathematical understanding is Brilliant.org, which provides interactive lessons and challenging problems across various mathematical disciplines, including trigonometry.