Solve over the Interval (2cos(theta)+1)(2sin(theta)+ square root of 3)=0 , [0,2pi)

Solution:

Step 1

Identify that the product of two terms equals zero, which implies that at least one of the terms must be zero.

$$2\cos(\theta) + 1 = 0 \quad \text{or} \quad 2\sin(\theta) + \sqrt{3} = 0$$

Step 2

Isolate $\cos(\theta)$ by setting $2\cos(\theta) + 1$ equal to zero and solving for $\theta$.

Step 2.1

Begin by setting the cosine term equal to zero.

$$2\cos(\theta) + 1 = 0$$

Step 2.2

Subtract $1$ from both sides to isolate the cosine term.

$$2\cos(\theta) = -1$$

Step 2.2.1

Divide by $2$ to solve for $\cos(\theta)$.

$$\cos(\theta) = -\frac{1}{2}$$

Step 2.2.2

Find the values of $\theta$ by taking the inverse cosine of $-\frac{1}{2}$.

$$\theta = \arccos\left(-\frac{1}{2}\right)$$

Step 2.2.3

Determine the general solution using the cosine function's periodicity.

$$\theta = \frac{2\pi}{3} + 2\pi n \quad \text{and} \quad \theta = \frac{4\pi}{3} + 2\pi n, \text{ for any integer } n$$

Step 3

Isolate $\sin(\theta)$ by setting $2\sin(\theta) + \sqrt{3}$ equal to zero and solving for $\theta$.

Step 3.1

Begin by setting the sine term equal to zero.

$$2\sin(\theta) + \sqrt{3} = 0$$

Step 3.2

Subtract $\sqrt{3}$ from both sides to isolate the sine term.

$$2\sin(\theta) = -\sqrt{3}$$

Step 3.2.1

Divide by $2$ to solve for $\sin(\theta)$.

$$\sin(\theta) = -\frac{\sqrt{3}}{2}$$

Step 3.2.2

Find the values of $\theta$ by taking the inverse sine of $-\frac{\sqrt{3}}{2}$.

$$\theta = \arcsin\left(-\frac{\sqrt{3}}{2}\right)$$

Step 3.2.3

Determine the general solution using the sine function's periodicity.

$$\theta = \frac{4\pi}{3} + 2\pi n \quad \text{and} \quad \theta = \frac{5\pi}{3} + 2\pi n, \text{ for any integer } n$$

Step 4

Combine the solutions from steps 2 and 3 to form the complete set of solutions.

$$\theta = \frac{2\pi}{3} + 2\pi n, \quad \theta = \frac{4\pi}{3} + 2\pi n, \quad \theta = \frac{5\pi}{3} + 2\pi n, \text{ for any integer } n$$

Step 5

Simplify the solutions by combining like terms.

$$\theta = \frac{2\pi}{3} + \pi n, \quad \theta = \frac{4\pi}{3} + 2\pi n, \text{ for any integer } n$$

Step 6

Identify the solutions that lie within the given interval $[0, 2\pi)$.

$$\theta = \frac{2\pi}{3}, \quad \theta = \frac{4\pi}{3}, \quad \theta = \frac{5\pi}{3}$$

Knowledge Notes:

1. Zero Product Property: If the product of two factors is zero, then at least one of the factors must be zero.

2. Trigonometric Functions: The cosine and sine functions have a range of $[-1, 1]$ and are periodic with a period of $2\pi$.

3. Inverse Trigonometric Functions: The $\arccos$ and $\arcsin$ functions are used to find the angle corresponding to a given cosine or sine value.

4. Periodicity of Trigonometric Functions: The general solutions for trigonometric equations involve adding integer multiples of the period ($2\pi$ for sine and cosine) to the specific solutions found within one period.

5. Quadrants and Signs: The sign of the sine and cosine functions depends on the quadrant in which the angle lies. Cosine is negative in the second and third quadrants, while sine is negative in the third and fourth quadrants.

6. Interval Notation: The notation $[0, 2\pi)$ represents the set of all real numbers from $0$ to $2\pi$, including $0$ but excluding $2\pi$.