Questions: Solve the equation. (Enter your answers as a comma-separated list. Use n as an arbitrary integer. Enter your response in radians.) x=4 sin ^2(x)+6 sin (x)+2=0

Solution Steps

To solve the given trigonometric equation, we can follow these steps:

1. Let \( y = \sin(x) \). This transforms the equation into a quadratic form: \( 4y^2 + 6y + 2 = 0 \).
2. Solve the quadratic equation for \( y \).
3. Find the corresponding \( x \) values by taking the inverse sine (arcsin) of the solutions for \( y \).
4. Since the sine function is periodic, include the general solution using \( n \) as an arbitrary integer.

We start with the equation: \[ x = 4 \sin^2(x) + 6 \sin(x) + 2 = 0 \] Letting \( y = \sin(x) \), we rewrite the equation as: \[ 4y^2 + 6y + 2 = 0 \]

Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we find the roots:

• Coefficients: \( a = 4, b = 6, c = 2 \)
• Discriminant: \( b^2 - 4ac = 6^2 - 4 \cdot 4 \cdot 2 = 36 - 32 = 4 \)
• Roots: \[ y_1 = \frac{-6 + \sqrt{4}}{2 \cdot 4} = \frac{-6 + 2}{8} = -\frac{1}{2} \] \[ y_2 = \frac{-6 - \sqrt{4}}{2 \cdot 4} = \frac{-6 - 2}{8} = -1 \]

Next, we find \( x \) values for the valid \( y \) values:

1. For \( y_1 = -\frac{1}{2} \): \[ x = \arcsin\left(-\frac{1}{2}\right) = -\frac{\pi}{6} \quad \text{and} \quad x = \pi - \left(-\frac{\pi}{6}\right) = \frac{7\pi}{6} \]
2. For \( y_2 = -1 \): \[ x = \arcsin(-1) = -\frac{\pi}{2} \quad \text{and} \quad x = \pi - \left(-\frac{\pi}{2}\right) = \frac{3\pi}{2} \]

The general solutions for \( x \) can be expressed as: \[ x = -\frac{\pi}{6} + 2n\pi, \quad x = \frac{7\pi}{6} + 2n\pi, \quad x = -\frac{\pi}{2} + 2n\pi, \quad x = \frac{3\pi}{2} + 2n\pi \] where \( n \) is any integer.

Final Answer