Problem 43 - Precalculus Essentials

Question

Involve trigonometric equations quadratic in form. Solve each equation on the interval \([0,2 \pi)\) $$2 \sin ^{2} x=\sin x+3$$

Step-by-Step Solution

Answer

The only solution within the interval \([0,2\pi)\) is \(x=\pi\).

Step 1: Make the equation in standard quadratic form

First, the equation will be rearranged into a standard quadratic form \(ax^2 + bx + c = 0\). This yields \(2 \sin^2x - \sin x - 3 = 0\)

Step 2: Apply the quadratic formula

Once in quadratic form, apply the quadratic formula \(\sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 2\), \(b = -1\), and \(c = -3\). This gives \(\sin x = \frac{1 \pm \sqrt{1 + 24}}{4} = \frac{1 \pm \sqrt{25}}{4} = \frac{1 \pm 5}{4}\). So, \(\sin x = 1.5\) or \(\sin x = -1\).

Step 3: Solve for all possible x values in given interval

Since the range of the sine function is \([-1, 1]\), \(\sin x = 1.5\) has no solution as it falls outside the range. \(\sin x = -1\) has a solution at \(\pi\). Therefore, the only solution in the interval \([0,2\pi)\) for the given equation is \(\pi\).

Key Concepts

Quadratic in Form Trigonometric Equations

Quadratic in form trigonometric equations involve terms where the trigonometric function is squared, resembling the structure of classic quadratic equations like those involving an unknown variable, usually expressed as 'x'. However, instead of the variable 'x', these equations utilize trigonometric functions such as sine (sin), cosine (cos), or tangent (tan).

For instance, an equation like \(2 \text{{sin}}^2 x = \text{{sin}} x + 3\) is quadratic in form because the squared sine term is analogous to an \(x^2\) term in a typical quadratic equation. To solve it, we transform the trigonometric equation into a standard quadratic form \(ax^2 + bx + c = 0\). This allows us to use familiar algebraic techniques to find the solutions.

Solving Trigonometric Equations

To solve trigonometric equations, particularly those that are quadratic in form, we utilize algebraic methods tailored for trigonometric functions. The first step in solving such an equation is to rewrite it in a standard quadratic form. Once this is achieved, we can apply algebraic methods like factoring, the quadratic formula, or even graphical methods to find the solutions.

When dealing with the standard form \(ax^2 + bx + c = 0\), where the variable 'x' is replaced with a trigonometric function, identifying possible solutions involves finding values for which the trigonometric function satisfies the equation. This often involves using inverse trigonometric functions to retrieve the angle measurements corresponding to the solutions.

Quadratic Formula

The quadratic formula is vital for solving equations that are quadratic in form, including trigonometric equations. This formula states that for any quadratic equation given as \(ax^2 + bx + c = 0\), the solutions can be found using \[x = \frac{{-b \text{{pm}} \sqrt{{b^2 - 4ac}}}}{{2a}}\].

Applying this to a trigonometric equation, the 'x' is replaced with the trigonometric function. For example, if we have \(2 \text{{sin}}^2 x - \text{{sin}} x - 3 = 0\), by substituting \(a = 2\), \(b = -1\), and \(c = -3\) into the quadratic formula, we can uncover the values of \(\text{{sin}} x\) that satisfy the equation. It's crucial to check if the calculated values are within the range of the trigonometric function involved.

Interval for Trigonometric Solutions

Trigonometric functions have specific ranges and periods that should be considered when finding solutions. For instance, the sine function has a range from -1 to 1, which means that any solutions resulting from an equation involving sine must fall within this interval. Moreover, when solving an equation over a specific interval, such as \([0, 2\pi)\), it's necessary to only consider the solutions that lie within that interval.

After applying the quadratic formula to a trigonometric equation, we obtain solutions for the trigonometric function that we must then translate back into angle measures. This step involves using inverse trigonometric functions or analyzing the unit circle to determine angles that correspond to the solutions—and it's crucial to include only angles that fall within the specified interval. For the equation \(2 \text{{sin}}^2 x = \text{{sin}} x + 3\), once we find that \(\text{{sin}} x = -1\), we identify \(\pi\) as the angle that satisfies the equation in the given interval.