Question

Statement 1 Number of solutions of the equation \(2 \sin ^{2} \theta-3 \sin \theta+1=0\) in the interval \([-2 \pi, 2 \pi]\) is 6 and Statement 2 General solution of \(\sin \theta=\mathrm{k}\) is given by \(\theta=\mathrm{n} \pi+(-1)^{\mathrm{n}} \alpha\) where \(\mathrm{n}\) is an integer and \(\sin \alpha=\mathrm{k}, 0 \leq \alpha<\pi\).

Short answer

Answer: There are 6 solutions in the given interval.

Step-by-step solution

Solve the equation 2sin^2θ - 3sinθ + 1 = 0

Since the given equation looks like a quadratic equation, let us make a substitution: Let \(x=\sin\theta\). Then our equation becomes \(2x^2 - 3x + 1 = 0\). Now, we can find the roots of this quadratic equation.

Find the roots of the quadratic equation

To find the roots of \(2x^2 - 3x + 1 = 0\), we can use the quadratic formula. The quadratic formula for a quadratic equation \(ax^2 + bx + c = 0\) is: \(x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\) Applying this to our equation, we get: \(x = \frac{3\pm\sqrt{(-3)^2-4(2)(1)}}{2(2)} = \frac{3\pm\sqrt{1}}{4}\) So, we have two roots: \(x_1 = \frac{3+\sqrt{1}}{4} = 1\) \(x_2 = \frac{3-\sqrt{1}}{4} = \frac{1}{2}\) These roots corresponds to the sine of the angles we are looking for, i.e. \(\sin\theta_1 = 1\) and \(\sin\theta_2 = \frac{1}{2}\).

Use the general solution of sinθ = k to find the angles

We are given the general solution of \(\sin\theta=k\) to be \(\theta=n\pi + (-1)^n\alpha\), where \(n\) is an integer and \(\sin\alpha=k\). We will use this formula to find the angles \(\theta_1\) and \(\theta_2\) for our two sine values \(x_1 = 1\) and \(x_2 = \frac{1}{2}\). For \(\sin\theta_1 = 1\), we have \(k=1\), and therefore \(\alpha = \arcsin(1) = \frac{\pi}{2}\). Using our general formula, we get \(\theta_1=n\pi + (-1)^n\frac{\pi}{2}\). For \(\sin\theta_2 = \frac{1}{2}\), we have \(k=\frac{1}{2}\), and therefore \(\alpha = \arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6}\). Using our general formula, we get \(\theta_2=n\pi + (-1)^n\frac{\pi}{6}\).

Find the specific solutions in the interval [-2π, 2π]

Now we need to find the values of \(n\) for which the specific solutions of \(\theta_1\) and \(\theta_2\) lie within the interval \([-2\pi, 2\pi]\). For \(\theta_1=n\pi+(-1)^n\frac{\pi}{2}\), we can try the following integer values of \(n\) in the given interval: - \(n = -2\) gives \(\theta_1 = -\frac{3\pi}{2}\) - \(n = -1\) gives \(\theta_1 = -\frac{\pi}{2}\) - \(n = 0\) gives \(\theta_1 = \frac{\pi}{2}\) - \(n = 1\) gives \(\theta_1 = \frac{3\pi}{2}\) For \(\theta_2=n\pi+(-1)^n\frac{\pi}{6}\), we can try the following integer values of \(n\) in the given interval: - \(n = -2\) gives \(\theta_2 = -\frac{11\pi}{6}\) - \(n = -1\) gives \(\theta_2 = -\frac{5\pi}{6}\) - \(n = 0\) gives \(\theta_2 = \frac{\pi}{6}\) - \(n = 1\) gives \(\theta_2 = \frac{5\pi}{6}\) Counting these solutions, we have 6 specific solutions in the interval \([-2\pi, 2\pi]\). Therefore, Statement 1 is true.

Key concepts

Quadratic Equations

Quadratic equations are pivotal in solving many mathematical problems and appear frequently in algebra. The equation is generally of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). These equations can be solved using various methods such as factoring, completing the square, or by using the quadratic formula.
The quadratic formula offers a straightforward way to find the roots of any quadratic equation. It is expressed as:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

This formula helps identify the values of \(x\) where the equation equals zero, and these values are the roots of the quadratic equation. In the case of the trigonometric equation \(2\sin^2\theta - 3\sin\theta + 1 = 0\), substituting \(\sin\theta\) with a variable \(x\) transformed it into the recognizable quadratic form, allowing us to easily solve for \(x\). By applying the quadratic formula, we determine the sine values that satisfy the given equation.
Understanding how to manipulate and solve quadratic equations is crucial, especially when dealing with trigonometric functions that resemble quadratic forms. This skill further assists in solving more complex mathematical problems and finding specific angle solutions.

General Solution of Trigonometric Functions

When discussing trigonometric equations, it's essential to comprehend the concept of their general solution. This provides a set of all possible solutions for a trigonometric equation. Specifically, for the sine function, the general solution is defined as \(\sin\theta = k\) and can be expressed in the formula \(\theta = n\pi + (-1)^n\alpha\), where:

• \(n\) represents an integer, indicating any whole number for possible solutions.
• \(\alpha = \arcsin(k)\), which implies finding the angle whose sine is equal to \(k\).
• The term \((-1)^n\) ensures the solutions account for both positive and negative orientations.

This comprehensive approach allows us to determine angles that satisfy specific trigonometric conditions. Using the general solution helps in finding a pattern or formula for any possible angle. In this particular example, the general solution is used to compute the angles corresponding to the roots obtained from the quadratic equation \(\sin\theta = 1\) and \(\sin\theta = \frac{1}{2}\).
These general solutions cover all angles, including those that repeat periodically on the unit circle, offering an infinite set of potential solutions, which can then be narrowed down based on the interval of interest.

Interval Solutions

Interval solutions are focused on finding solutions to an equation within a specified range, or interval, of values. This is particularly significant in trigonometric equations where solutions can be infinite due to the periodic nature of trigonometric functions. For the equation \(2\sin^2\theta - 3\sin\theta + 1 = 0\), we need to find solutions specifically within the interval \([-2\pi, 2\pi]\).
This step involves using the general solutions of the derived equations and identifying which angles fit within the given interval. Each value of \(n\) from the general solution will yield an angle \(\theta\) that can be checked against our interval:

• For \(\sin\theta = 1\), solutions \(\theta = n\pi + (-1)^n\frac{\pi}{2}\) are determined for different \(n\).
• For \(\sin\theta = \frac{1}{2}\), solutions \(\theta = n\pi + (-1)^n\frac{\pi}{6}\) are calculated similarly.

We determine the integer values of \(n\) ensuring angle solutions fall within \([-2\pi, 2\pi]\). Counting these values gives us the total number of solutions, which, in this case, validates the statement that there are six solutions. Grasping interval solutions aid in solving real-world trigonometric problems where solutions must lie within specific boundaries.