you know that sin π/4 = √2/2
sin(x) is negative in QIII and QIV
the rest should be easy.
sin(x) is negative in QIII and QIV
the rest should be easy.
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.
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.