Question

The number of values of \(x\) in the interval \(\left[0, \frac{11 \pi}{2}\right]\) satisfying the equation \(6 \sin ^{2} x+\sin\) \(x-2=0\) is (1) 9 (2) 10 (3) 11 (4) 12

Short answer

The correct answer is not listed among the choices. Total number of solutions is 15.

Step-by-step solution

- Transform the Equation

The given equation is: \[6 \sin^2 x + \sin x - 2 = 0\]Let's substitute \( \sin x \) with a new variable \( y \). Thus, the equation transforms to: \[6y^2 + y - 2 = 0\]

- Solve the Quadratic Equation

Now, solve the quadratic equation \(6y^2 + y - 2 = 0\). To do this, we can use the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]with \(a = 6\), \(b = 1\), and \(c = -2\). Substituting the values, we get: \[ y = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 6 \cdot (-2)}}{2 \cdot 6} \]This simplifies to:\[ y = \frac{-1 \pm \sqrt{1 + 48}}{12} \]\[ y = \frac{-1 \pm 7}{12} \]Thus, the solutions are: \[y_1 = \frac{6}{12} = \frac{1}{2}\]\[y_2 = \frac{-8}{12} = -\frac{2}{3}\]

- Convert Back to Sine Terms

Recall the substitution, \(y = \sin x\). Therefore, we have: \[ \sin x = \frac{1}{2} \]\[ \sin x = -\frac{2}{3} \]

- Find Values of x for \( \sin x = \frac{1}{2} \)

For \( \sin x = \frac{1}{2} \): \[ x = \frac{\pi}{6}, \frac{5\pi}{6} \,\] plus multiples of \(2\pi\). In the interval \( \left[0, \frac{11\pi}{2}\right] \), the values are: \[\frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}, \frac{17\pi}{6}, \frac{25\pi}{6}, \frac{29\pi}{6} \]which gives 6 solutions.

- Find Values of x for \( \sin x = -\frac{2}{3} \)

For \( \sin x = -\frac{2}{3} \): Using the principal solution and its periodicity, we find: \[ x = \arcsin(-\frac{2}{3}) \ \text{ and } \ \pi - \arcsin(-\frac{2}{3}) \] plus multiples of \(2\pi\). In the interval \( \left[0, \frac{11\pi}{2}\right] \), the values are: \[\arcsin(-\frac{2}{3}), \pi + \arcsin(-\frac{2}{3}), 2\pi + \arcsin(-\frac{2}{3}), 3\pi + \arcsin(-\frac{2}{3}), 4\pi + \arcsin(-\frac{2}{3}), \] \[ \pi - \arcsin(-\frac{2}{3}), 2\pi - \arcsin(-\frac{2}{3}), 3\pi - \arcsin(-\frac{2}{3}), 4\pi - \arcsin(-\frac{2}{3}) \]which gives 9 solutions.

- Total Number of Solutions

Summing the solutions from both cases: 6 (for \( \sin x = \frac{1}{2} \)) + 9 (for \( \sin x = -\frac{2}{3} \)). This totals to 15 solutions.

Key concepts

Quadratic equations

Quadratic equations are polynomial equations of the form \( ax^2 + bx + c = 0 \). They are called 'quadratic' because the highest exponent of the variable is 2. A common way to solve quadratic equations is by using the quadratic formula: \[ x = \frac{-b \, \text{±} \, \sqrt{b^2 - 4ac}}{2a} \]This formula helps find the values of \( x \) that satisfy the equation. In the given problem, we first replace \( \sin x \) with \( y \), transforming the trigonometric equation to a quadratic form: \( 6y^2 + y - 2 = 0 \). Using the quadratic formula with \( a = 6 \), \( b = 1 \), and \( c = -2 \), we find two solutions for \( y \): \( y_1 = \frac{1}{2} \) and \( y_2 = -\frac{2}{3} \). These solutions are then translated back into terms involving \( \sin x \).

Sine function

The sine function, denoted as \( \sin(x) \), is a fundamental trigonometric function that represents the ratio of the opposite side to the hypotenuse in a right triangle. It varies between -1 and 1. In the context of our equation, we substituted \( \sin x \) with \( y \) and solved for \( y \).

• For \( \sin x = \frac{1}{2} \), we find that \( x \) can be \( \frac{\pi}{6} \), \( \frac{5\pi}{6} \), and repeating every \( 2\pi \).
• For \( \sin x = -\frac{2}{3} \), we use the inverse sine function \( \arcsin \) to find the principal solutions.

Both sets of solutions must be evaluated within the specified interval to find all possible values of \( x \) that satisfy the original trigonometric equation.

Interval solutions

Interval solutions involve determining which values comply with both the equation and the given interval. In our problem, the interval is \( [0, \frac{11\pi}{2}] \). After finding the general solutions for \( x \) from both \( \sin x = \frac{1}{2} \) and \( \sin x = -\frac{2}{3} \), we list all values within this interval. For instance, for \( \sin x = \frac{1}{2} \), we get solutions like \( \frac{\pi}{6} \), \( \frac{13\pi}{6} \), etc. Each solution must be adjusted by adding multiples of \( 2\pi \) until the upper limit \( \frac{11\pi}{2} \) is reached. This gives 6 solutions for \( \frac{1}{2} \).Similarly, we find 9 solutions for \( -\frac{2}{3} \). Adding these, we total 15 solutions. Thus, understanding interval solutions is crucial for solving these types of trigonometric equations.