Solve over the Interval sin(2x)=-1/2 , [0,pi]

Solution:

Step 1:

Isolate $x$ by applying the inverse sine function to both sides: $2x = \sin^{-1}\left(-\frac{1}{2}\right)$.

Step 2:

Resolve the inverse sine to its principal value: $2x = -\frac{\pi}{6}$.

Step 3:

Isolate $x$ by dividing the equation by 2: $x = \frac{-\frac{\pi}{6}}{2}$.

Step 4:

Simplify the fraction: $x = -\frac{\pi}{12}$.

Step 5:

Locate the second solution by using the sine function's periodicity. Since sine is negative in the third and fourth quadrants, add the reference angle to $\pi$: $2x = \pi + \frac{\pi}{6}$.

Step 6:

Simplify to find the second solution: $2x = \frac{7\pi}{6}$.

Step 7:

Divide by 2 to isolate $x$: $x = \frac{\frac{7\pi}{6}}{2}$.

Step 8:

Further simplify the fraction: $x = \frac{7\pi}{12}$.

Step 9:

Calculate the sine function's period with $b = 2$: Period $= \frac{2\pi}{|b|} = \frac{2\pi}{2} = \pi$.

Step 10:

Adjust negative solutions to fall within the given interval by adding $\pi$: $x = -\frac{\pi}{12} + \pi = \frac{11\pi}{12}$.

Step 11:

List all solutions within the interval $[0, \pi]$: $x = \frac{7\pi}{12}, \frac{11\pi}{12}$.

Step 12:

Include the periodicity to express the general solution: $x = \frac{7\pi}{12} + \pi n, \frac{11\pi}{12} + \pi n$ for any integer $n$.

Step 13:

Determine the values of $n$ that yield solutions within the interval $[0, \pi]$. For $n = 0$, we get $x = \frac{7\pi}{12}$ and $x = \frac{11\pi}{12}$, both of which are in the interval.

Final Answer:

$x = \frac{7\pi}{12}, \frac{11\pi}{12}$.

Knowledge Notes:

1. Inverse Trigonometric Functions: These functions are used to find the angle that produces a given trigonometric value. For example, $\sin^{-1}(-\frac{1}{2})$ gives the angle whose sine is $-\frac{1}{2}$.

2. Simplifying Fractions: This involves reducing the complexity of a fraction by dividing the numerator and denominator by their greatest common factor or by multiplying by the reciprocal if necessary.

3. Periodicity of Trigonometric Functions: Sine and cosine functions have a period of $2\pi$, meaning their values repeat every $2\pi$ radians. For functions like $\sin(2x)$, the period is $\frac{2\pi}{|b|}$, where $b$ is the coefficient of $x$ inside the function.

4. Quadrants: The sine function is positive in the first and second quadrants and negative in the third and fourth quadrants. This knowledge is used to find the correct angle that satisfies the original equation when the sine value is negative.

5. Interval Consideration: When solving trigonometric equations, it is important to consider the interval over which the solution is sought. Solutions outside the given interval may need to be adjusted using the function's periodicity to fall within the interval.

6. General Solution for Trigonometric Equations: The general solution accounts for the periodic nature of trigonometric functions and is expressed as $x = x_0 + nP$, where $x_0$ is a particular solution, $P$ is the period, and $n$ is an integer.