Solve Cos ⁡ 2 ( X ) − Cos ⁡ ( X ) = 0 \cos^2(x) - \cos(x) = 0 Cos 2 ( X ) − Cos ( X ) = 0 For X X X , Where 0 ≤ X ≤ 2 Π 0 \leq X \leq 2\pi 0 ≤ X ≤ 2 Π . Select All That Apply:- 2 Π 2\pi 2 Π - 0 0 0 - Π 2 \frac{\pi}{2} 2 Π ​ - 3 Π 2 \frac{3\pi}{2} 2 3 Π ​ - Π \pi Π

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Introduction

In this article, we will delve into solving the trigonometric equation $\cos^2(x) - \cos(x) = 0$ for $x$, where $0 \leq x \leq 2\pi$. This equation involves the cosine function, and we will use various trigonometric identities and techniques to find the solutions for $x$.

Understanding the Equation

The given equation is $\cos^2(x) - \cos(x) = 0$. To solve this equation, we can start by factoring out the common term $\cos(x)$ from both terms. This gives us:

$$\cos(x)(\cos(x) - 1) = 0$$

Factoring and Solving

Now, we can set each factor equal to zero and solve for $x$. This gives us two separate equations:

$$\cos(x) = 0 \quad \text{or} \quad \cos(x) - 1 = 0$$

Solving $\cos(x) = 0$

To solve the equation $\cos(x) = 0$, we need to find the values of $x$ for which the cosine function is equal to zero. The cosine function is equal to zero at odd multiples of $\frac{\pi}{2}$. Therefore, we have:

$$x = \frac{(2k + 1)\pi}{2}$$

where $k$ is an integer.

Solving $\cos(x) - 1 = 0$

To solve the equation $\cos(x) - 1 = 0$, we can add 1 to both sides of the equation, which gives us:

$$\cos(x) = 1$$

The cosine function is equal to 1 at multiples of $2\pi$. Therefore, we have:

$$x = 2k\pi$$

where $k$ is an integer.

Finding the Solutions for $x$

Now that we have found the solutions for both equations, we can combine them to find the solutions for the original equation. We have:

$$x = \frac{(2k + 1)\pi}{2} \quad \text{or} \quad x = 2k\pi$$

Evaluating the Solutions

We need to evaluate the solutions for $x$ in the given interval $0 \leq x \leq 2\pi$. We can plug in the values of $k$ to find the corresponding values of $x$.

For the first equation, we have:

$$x = \frac{(2k + 1)\pi}{2}$$

Plugging in $k = 0$, we get:

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

Plugging in $k = 1$, we get:

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

Plugging in $k = 2$, we get:

$$x = \frac{5\pi}{2}$$

However, this value of $x$ is outside the given interval $0 \leq x \leq 2\pi$. Therefore, we can ignore this solution.

For the second equation, we have:

$$x = 2k\pi$$

Plugging in $k = 0$, we get:

$$x = 0$$

Plugging in $k = 1$, we get:

$$x = 2\pi$$

Conclusion

In conclusion, the solutions for the equation $\cos^2(x) - \cos(x) = 0$ for $x$ in the interval $0 \leq x \leq 2\pi$ are:

• $\frac{\pi}{2}$
• $\frac{3\pi}{2}$
• $0$
• $2\pi$

Therefore, the correct answers are:

• $\frac{\pi}{2}$
• $\frac{3\pi}{2}$
• $0$
• $2\pi$
  

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Introduction

In our previous article, we solved the trigonometric equation $\cos^2(x) - \cos(x) = 0$ for $x$, where $0 \leq x \leq 2\pi$. We found the solutions for $x$ and evaluated them in the given interval. In this article, we will provide a Q&A section to help clarify any doubts or questions that readers may have.

Q: What is the main difference between the two equations $\cos(x) = 0$ and $\cos(x) - 1 = 0$?

A: The main difference between the two equations is the value of the cosine function. In the equation $\cos(x) = 0$, the cosine function is equal to zero, which occurs at odd multiples of $\frac{\pi}{2}$. In the equation $\cos(x) - 1 = 0$, the cosine function is equal to 1, which occurs at multiples of $2\pi$.

Q: How do we find the solutions for $x$ in the equation $\cos(x) = 0$?

A: To find the solutions for $x$ in the equation $\cos(x) = 0$, we need to find the values of $x$ for which the cosine function is equal to zero. This occurs at odd multiples of $\frac{\pi}{2}$, which can be expressed as:

$$x = \frac{(2k + 1)\pi}{2}$$

where $k$ is an integer.

Q: How do we find the solutions for $x$ in the equation $\cos(x) - 1 = 0$?

A: To find the solutions for $x$ in the equation $\cos(x) - 1 = 0$, we need to find the values of $x$ for which the cosine function is equal to 1. This occurs at multiples of $2\pi$, which can be expressed as:

$$x = 2k\pi$$

where $k$ is an integer.

Q: What is the significance of the interval $0 \leq x \leq 2\pi$ in this problem?

A: The interval $0 \leq x \leq 2\pi$ is significant because it restricts the possible values of $x$ to a specific range. This allows us to evaluate the solutions for $x$ and determine which ones are valid.

Q: How do we evaluate the solutions for $x$ in the given interval?

A: To evaluate the solutions for $x$ in the given interval, we need to plug in the values of $k$ into the equations and determine which ones fall within the interval $0 \leq x \leq 2\pi$.

Q: What are the final solutions for $x$ in the equation $\cos^2(x) - \cos(x) = 0$ for $x$ in the interval $0 \leq x \leq 2\pi$?

A: The final solutions for $x$ in the equation $\cos^2(x) - \cos(x) = 0$ for $x$ in the interval $0 \leq x \leq 2\pi$ are:

• $\frac{\pi}{2}$
• $\frac{3\pi}{2}$
• $0$
• $2\pi$

Conclusion

In conclusion, the Q&A section provides additional clarification and insight into the problem of solving the trigonometric equation $\cos^2(x) - \cos(x) = 0$ for $x$ in the interval $0 \leq x \leq 2\pi$. We hope that this article has been helpful in understanding the problem and its solutions.