c)2sin^2(x)+cos(x)-1=0 
d)tan(x)+cot(x)=-4

solve x in the domain of [0,2$$ \pi] $$

Let's solve each equation for $ x $ in the domain $ [0, 2\pi] $.

### Part (c): Solve $ 2\sin^2(x) + \cos(x) - 1 = 0 $

1. **Express $\sin^2(x)$ using $\cos(x)$:**
   We know that $ \sin^2(x) = 1 - \cos^2(x) $. Substitute this into the equation:
   \[
   2(1 - \cos^2(x)) + \cos(x) - 1 = 0
   \]
   Simplify:
   \[
   2 - 2\cos^2(x) + \cos(x) - 1 = 0
   \]
   \[
   -2\cos^2(x) + \cos(x) + 1 = 0
   \]
   Multiply through by $-1$ to simplify:
   \[
   2\cos^2(x) - \cos(x) - 1 = 0
   \]

2. **Solve the quadratic equation:**
   This is a quadratic equation in terms of $\cos(x)$. Let $ y = \cos(x) $, so the equation becomes:
   \[
   2y^2 - y - 1 = 0
   \]
   Factor the quadratic:
   \[
   (2y + 1)(y - 1) = 0
   \]
   So, $ y = -\frac{1}{2} $ or $ y = 1 $.

3. **Find $ x $ using the solutions for $ y $:**

   - For $ \cos(x) = -\frac{1}{2} $:
     \[
     x = \frac{2\pi}{3}, \frac{4\pi}{3}
     \]
   - For $ \cos(x) = 1 $:
     \[
     x = 0
     \]

   So, the solutions in the interval $ [0, 2\pi] $ are:
   \[
   x = 0, \frac{2\pi}{3}, \frac{4\pi}{3}
   \]

### Part (d): Solve $ \tan(x) + \cot(x) = -4 $

1. **Express $\cot(x)$ as $\frac{1}{\tan(x)}$:**
   \[
   \tan(x) + \frac{1}{\tan(x)} = -4
   \]
   Multiply both sides by $\tan(x)$ to eliminate the fraction:
   \[
   \tan^2(x) + 1 = -4\tan(x)
   \]
   Rearrange the equation:
   \[
   \tan^2(x) + 4\tan(x) + 1 = 0
   \]

2. **Solve the quadratic equation:**
   Let $ y = \tan(x) $, so the equation becomes:
   \[
   y^2 + 4y + 1 = 0
   \]
   Use the quadratic formula:
   \[
   y = \frac{-4 \pm \sqrt{4^2 - 4(1)(1)}}{2(1)} = \frac{-4 \pm \sqrt{16 - 4}}{2} = \frac{-4 \pm \sqrt{12}}{2} = \frac{-4 \pm 2\sqrt{3}}{2} = -2 \pm \sqrt{3}
   \]

3. **Find $ x $ using the solutions for $ y = \tan(x) $:**
   - For $ \tan(x) = -2 + \sqrt{3} $
   - For $ \tan(x) = -2 - \sqrt{3} $

   You'll need to find the values of $ x $ in the interval $ [0, 2\pi] $ that satisfy these conditions using the inverse tangent function. However, the exact forms are:
   \[
   x = \tan^{-1}(-2 + \sqrt{3}), \pi + \tan^{-1}(-2 + \sqrt{3})
   \]
   and
   \[
   x = \tan^{-1}(-2 - \sqrt{3}), \pi + \tan^{-1}(-2 - \sqrt{3})
   \]

Given that these values are not standard angles, they would need to be approximated or used directly in the trigonometric functions for further evaluation.

Would you like to explore any of these solutions further or need more details? Here are some related questions you might find interesting:

1. What are the exact values for $ x $ in part (d) using a calculator?
2. How do the solutions for $ \tan(x) + \cot(x) = -4 $ differ on the unit circle?
3. Can you graph the functions to visualize the solutions?
4. What happens if the equation in part (d) is changed to $ \tan(x) + \cot(x) = 4 $?
5. How do you verify the solutions on the interval $ [0, 2\pi] $?
6. What are the implications of having multiple solutions in trigonometric equations?
7. How would the solution process differ if the domain were $ [0, \pi] $ instead?
8. How do trigonometric identities simplify solving such equations?

**Tip:** When solving trigonometric equations, always consider the periodic nature of the functions to ensure all possible solutions within the given domain are found.

Learn how to solve trigonometric equations 2sin^2(x) + cos(x) - 1 = 0 and tan(x) + cot(x) = -4 step-by-step with detailed explanations. This problem involves using trigonometric identities and solving quadratic equations in the context of trigonometry. Suitable for advanced high school students and those preparing for college-level mathematics.

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