Find All Solutions Of The Equation In The Interval \[0, 2\pi$\].$\cos 2x + \sin X = 0$Write Your Answer In Radians In Terms Of $\pi$. If There Is More Than One Solution, Separate Them With Commas.$x = \square

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Introduction

Solving trigonometric equations can be a challenging task, especially when dealing with multiple trigonometric functions. In this article, we will focus on finding all solutions of the equation $\cos 2x + \sin x = 0$ in the interval $[0, 2\pi]$. We will use various trigonometric identities and techniques to simplify the equation and find its solutions.

Using Trigonometric Identities

To solve the equation $\cos 2x + \sin x = 0$, we can use the double-angle identity for cosine: $\cos 2x = 1 - 2\sin^2 x$. Substituting this into the equation, we get:

$$1 - 2\sin^2 x + \sin x = 0$$

Rearranging the Equation

We can rearrange the equation to isolate the sine term:

$$-2\sin^2 x + \sin x + 1 = 0$$

Factoring the Quadratic

The equation is a quadratic equation in terms of $\sin x$. We can factor it as:

$$(-2\sin x + 1)(\sin x + 1) = 0$$

Setting Each Factor Equal to Zero

To find the solutions, we set each factor equal to zero:

$$-2\sin x + 1 = 0 \quad \text{or} \quad \sin x + 1 = 0$$

Solving for $\sin x$

Solving for $\sin x$ in each equation, we get:

$$\sin x = \frac{1}{2} \quad \text{or} \quad \sin x = -1$$

Finding the Solutions

We can use the unit circle or a trigonometric table to find the solutions for $\sin x = \frac{1}{2}$ and $\sin x = -1$.

For $\sin x = \frac{1}{2}$, we have:

$$x = \frac{\pi}{6}, \frac{5\pi}{6}$$

For $\sin x = -1$, we have:

$$x = \frac{3\pi}{2}$$

Checking the Solutions

We need to check if the solutions we found are in the interval $[0, 2\pi]$. We can see that all three solutions are in the interval.

Conclusion

In this article, we used trigonometric identities and techniques to solve the equation $\cos 2x + \sin x = 0$ in the interval $[0, 2\pi]$. We found three solutions: $x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}$. These solutions are in radians and are expressed in terms of $\pi$.

Final Answer

The final answer is: $\boxed{\frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}}$

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Introduction

In our previous article, we solved the equation $\cos 2x + \sin x = 0$ in the interval $[0, 2\pi]$. We found three solutions: $x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}$. In this article, we will answer some common questions related to solving this equation.

Q: What is the first step in solving the equation $\cos 2x + \sin x = 0$?

A: The first step in solving the equation $\cos 2x + \sin x = 0$ is to use the double-angle identity for cosine: $\cos 2x = 1 - 2\sin^2 x$. Substituting this into the equation, we get: $1 - 2\sin^2 x + \sin x = 0$.

Q: How do I simplify the equation $1 - 2\sin^2 x + \sin x = 0$?

A: We can simplify the equation by rearranging it to isolate the sine term: $-2\sin^2 x + \sin x + 1 = 0$. This is a quadratic equation in terms of $\sin x$.

Q: How do I factor the quadratic equation $-2\sin^2 x + \sin x + 1 = 0$?

A: We can factor the quadratic equation as: $(-2\sin x + 1)(\sin x + 1) = 0$.

Q: What are the two possible solutions for the equation $-2\sin x + 1 = 0$?

A: The two possible solutions for the equation $-2\sin x + 1 = 0$ are: $\sin x = \frac{1}{2}$ and $\sin x = -1$.

Q: How do I find the solutions for $\sin x = \frac{1}{2}$ and $\sin x = -1$?

A: We can use the unit circle or a trigonometric table to find the solutions for $\sin x = \frac{1}{2}$ and $\sin x = -1$. For $\sin x = \frac{1}{2}$, we have: $x = \frac{\pi}{6}, \frac{5\pi}{6}$. For $\sin x = -1$, we have: $x = \frac{3\pi}{2}$.

Q: Are the solutions $x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}$ in the interval $[0, 2\pi]$?

A: Yes, the solutions $x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}$ are in the interval $[0, 2\pi]$.

Q: What is the final answer to the equation $\cos 2x + \sin x = 0$?

A: The final answer to the equation $\cos 2x + \sin x = 0$ is: $\boxed{\frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}}$.

Conclusion

In this article, we answered some common questions related to solving the equation $\cos 2x + \sin x = 0$. We hope that this article has been helpful in clarifying any doubts you may have had about solving this equation.

Final Tips

• Always use the double-angle identity for cosine when solving equations involving $\cos 2x$.
• Rearrange the equation to isolate the sine term.
• Factor the quadratic equation to find the possible solutions.
• Use the unit circle or a trigonometric table to find the solutions for $\sin x = \frac{1}{2}$ and $\sin x = -1$.
• Check if the solutions are in the interval $[0, 2\pi]$.

Related Articles

• Solving the Equation $\cos 2x + \sin x = 0$ in the Interval $[0, 2\pi]$
• Trigonometric Identities and Formulas
• Solving Quadratic Equations in Terms of $\sin x$