Find the exact values of x in the interval [0, 4π] that satisfy the equation sin x = -√2 / 2 (refer to y = sin x or y = cos x )

To find the exact values of x that satisfy the equation sin(x) = -√2/2 in the interval [0, 4π], we need to refer to the unit circle and the values of sin and cos for various angles.

1. Start by drawing the unit circle, which is a circle with a radius of 1 centered at the origin (0, 0) on the Cartesian plane.

2. Mark the points on the unit circle where the y-coordinate is -√2/2, which corresponds to the equation sin(x) = -√2/2. These points are located on the unit circle at the angles π/4 and 7π/4.

3. Identify the values of x that correspond to these points on the unit circle. For sin(x) = -√2/2, the angles π/4 and 7π/4 satisfy the equation.

4. However, we need to consider the interval [0, 4π]. Since 7π/4 is greater than 4π, it is outside the specified interval and is not a valid solution.

5. Therefore, the only exact value of x that satisfies the equation sin(x) = -√2/2 in the interval [0, 4π] is x = π/4.

To summarize, the exact value of x in the interval [0, 4π] that satisfies the equation sin(x) = -√2/2 is x = π/4.

To find the exact values of x in the interval [0, 4π] that satisfy the equation sin x = -√2 / 2, we can refer to the unit circle and the special angles for sine function.

The angle whose sine value is -√2/2 is -π/4, which can be found at the coordinate (1/√2, -1/√2) on the unit circle.

In the interval [0, 4π], we have the following special angles for sine function:
0, π/6, π/4, π/3, π/2, 2π/3, 3π/4, 5π/6, π, 7π/6, 5π/4, 4π/3, 3π/2, 5π/3, 7π/4, 11π/6, 2π.

We can see that there are four solutions in the given interval where sin x = -√2/2:
x = -π/4, 7π/4, 5π/4, and 11π/4.

you know that sin π/4 = √2/2

sin(x) is negative in QIII and QIV
the rest should be easy.