Questions: Solve, finding all solutions in [0,2 pi) or [0°, 360°). - sin 2 x cos x + sin x = 0 After applying the appropriate double angle-identity and simplifying, write the completely factored form of the equation. - sin x(2 cos^2 x - 1) = 0 (Type your answer in factored form.) The solution(s) in [0°, 360°) is/are x = . (Type your answer in degrees. Do not include the degree symbol in your answer. Use a comma to separate answers as needed.)

Solve, finding all solutions in [0,2 pi) or [0°, 360°).
- sin 2 x cos x + sin x = 0

After applying the appropriate double angle-identity and simplifying, write the completely factored form of the equation.
- sin x(2 cos^2 x - 1) = 0
(Type your answer in factored form.)
The solution(s) in [0°, 360°) is/are x = .
(Type your answer in degrees. Do not include the degree symbol in your answer. Use a comma to separate answers as needed.)

Transcript text: Solve, finding all solutions in $[0,2 \pi)$ or $\left[0^{\circ}, 360^{\circ}\right)$. \[ -\boldsymbol{\operatorname { s i n }} 2 \mathrm{x} \boldsymbol{\operatorname { c o s }} \mathrm{x}+\boldsymbol{\operatorname { s i n }} \mathrm{x}=0 \] After applying the appropriate double angle-identity and simplifying, write the completely factored form of the equation. \[ -\sin x\left(2 \cos ^{2} x-1\right)=0 \] (Type your answer in factored form.) The solution(s) in $\left[0^{\circ}, 360^{\circ}\right)$ is/are $\mathrm{x}=$ $\square$ . (Type your answer in degrees. Do not include the degree symbol in your answer. Use a comma to separate answers as needed.)

Solution

Solution Steps

To solve the given trigonometric equation, we first apply the double angle identity for sine, which is $\sin(2x) = 2\sin(x)\cos(x)$. This allows us to rewrite the equation in a simpler form. Next, we factor the equation to find the solutions for $x$ within the specified interval. We solve each factor separately to find all possible solutions in degrees.

Step 1: Rewrite the Equation

We start with the equation: \[ -\sin(2x) \cos(x) + \sin(x) = 0 \] Using the double angle identity $\sin(2x) = 2\sin(x)\cos(x)$, we can rewrite the equation as: \[ -\sin(x)(2\cos^2(x) - 1) = 0 \]

Step 2: Factor the Equation

The equation can be factored into two parts: \[ -\sin(x) = 0 \quad \text{or} \quad 2\cos^2(x) - 1 = 0 \]

Step 3: Solve Each Factor

For $-\sin(x) = 0$: \[ \sin(x) = 0 \implies x = 0, 360 \text{ degrees} \]
For $2\cos^2(x) - 1 = 0$: \[ 2\cos^2(x) = 1 \implies \cos^2(x) = \frac{1}{2} \implies \cos(x) = \pm \frac{1}{\sqrt{2}} \implies x = 45, 135, 225, 315 \text{ degrees} \]

Step 4: Compile All Solutions

Combining the solutions from both factors, we have: \[ x = 0, 45, 135, 225, 315, 360 \text{ degrees} \]