Question

How many values of x in [0, 2pi] satisfy sin(2x) = 0?

Answer 1

Final Answer:

There are two values of \(x\) in \([0, 2\pi]\) that satisfy \(\sin(2x) = 0\).

Step-by-step explanation:

In the given equation \(\sin(2x) = 0\), we are looking for values of \(x\) that make the sine of twice that value equal to zero. The sine function is zero at multiples of \(\pi\), so we set \(2x\) equal to \(n\pi\), where \(n\) is an integer. Solving for \(x\), we get \(x = \frac{n\pi}{2}\).

Since we are looking for solutions in the interval \([0, 2\pi]\), we consider values of \(n\) such that \(0 \leq \frac{n\pi}{2} \leq 2\pi\). The solutions occur when \(n\) is even, leading to \(x = 0\) and \(x = \pi\), satisfying the given equation. Thus, there are two values of \(x\) in the specified interval that satisfy \(\sin(2x) = 0\).

In summary, by leveraging the properties of the sine function and the specified interval, we determined that the solutions to \(\sin(2x) = 0\) in \([0, 2\pi]\) are \(x = 0\) and \(x = \pi\). These values align with the points where the sine function equals zero due to its periodic nature.

Answer 2

There are two values of x in the interval [0, 2π] that satisfy sin(2x) = 0.

Step-by-step explanation:

In the given interval, [0, 2π], the equation sin(2x) = 0 suggests that the sine of 2x is zero. To find the values of x that satisfy this equation, we consider where the sine function equals zero within this range. The sine function equals zero at specific angles in the unit circle, which are 0, π, 2π, and so on. However, in this case, we're dealing with the double angle, 2x, meaning we're looking for angles where sin(2x) = 0.

The solutions for sin(2x) = 0 occur when the angle 2x is such that sin(2x) = 0. Considering the unit circle, sin(2x) = 0 when 2x equals 0 radians, π radians, 2π radians, and so forth. To obtain the values for x within the interval [0, 2π], we divide these angles by 2, resulting in x = 0/2, π/2, 2π/2, which simplifies to x = 0, π/2. These values satisfy the given equation sin(2x) = 0 within the specified interval.

Therefore, after analyzing the relationship between sin(2x) and the unit circle within the given interval [0, 2π], we find that there are precisely two values of x, namely 0 and π/2, that satisfy the equation sin(2x) = 0.