Questions: Solve the equation for exact solutions over the interval [0 ; 2π). 2 sin^2 x+4 sin x+2=0 Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The solution set is . (Type an exact answer, using π as needed. Type your answer in radians. Use integers or fractions for any numbers in the expression. Use a comma to separate answers as needed.) B. The solution is the empty set.

$Solve the equation for exact solutions over the interval [0 ; 2π). 2 sin^2 x+4 sin x+2=0 Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The solution set is . (Type an exact answer, using π as needed. Type your answer in radians. Use integers or fractions for any numbers in the expression. Use a comma to separate answers as needed.) B. The solution is the empty set.$

Transcript text: Solve the equation for exact solutions over the interval $[0 ; 2 \pi)$. \[ 2 \sin ^{2} x+4 \sin x+2=0 \] Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The solution set is $\square$ \}. (Type an exact answer, using $\pi$ as needed. Type your answer in radians. Use integers or fractions for any numbers in the expression. Use a comma to separate answers as needed.) B. The solution is the empty set.

Solution

Solution Steps

To solve the equation $2 \sin^2 x + 4 \sin x + 2 = 0$ over the interval $[0, 2\pi)$, we can follow these steps:

Recognize that this is a quadratic equation in terms of $\sin x$.
Let $u = \sin x$, then the equation becomes $2u^2 + 4u + 2 = 0$.
Solve the quadratic equation for $u$.
Check the solutions to see if they fall within the range of $\sin x$, which is $[-1, 1]$.
Convert the valid solutions for $u$ back to $x$ using the inverse sine function and find all solutions within the given interval.

Step 1: Identify the Quadratic Equation

We start with the equation: \[ 2 \sin^2 x + 4 \sin x + 2 = 0 \] This can be rewritten as a standard quadratic equation in terms of $u = \sin x$: \[ 2u^2 + 4u + 2 = 0 \]

Step 2: Calculate the Discriminant

The discriminant $D$ of the quadratic equation is calculated as follows: \[ D = b^2 - 4ac = 4^2 - 4 \cdot 2 \cdot 2 = 16 - 16 = 0 \] Since the discriminant is $0$, there is exactly one real solution for $u$.

Step 3: Solve for $u$

Using the quadratic formula: \[ u = \frac{-b \pm \sqrt{D}}{2a} = \frac{-4 \pm 0}{2 \cdot 2} = \frac{-4}{4} = -1 \] Thus, we have: \[ \sin x = -1 \]

Step 4: Find $x$ in the Interval $[0, 2\pi)$

The solution for $\sin x = -1$ occurs at: \[ x = \frac{3\pi}{2} \] This value is within the interval $[0, 2\pi)$.