Solve the equation in the interval from 510^\circ510 ∘ 510, degrees to 1050^\circ1050 ∘ 1050, degrees. Your answer should be in degrees. \sin(x)=0sin(x)=0

Step-by-step explanation:

Given equation sin(x) = 0 lying between the interval 510°≤x≤1050°. To find the value of x within this range, we first find the value of from the equation as shown;

sin(x) = 0

taking the arcsin of both sides;

Since x is positive in the first and second quadrant, in the second quadrant, the value of x will be 180°-0° = 180°

Subsequent value of x equivalent to 0° will be addition of 180° to each previous value gotten. The values of x within the range given are as shown

x1 =0°

x2 = 180°-0° = 180(2nd quadrant)

x3 = 180°+180° = 360°

x4 = 360°+180° = 540°

x5 = 540°+180°=720°

x6 = 720°+180° = 900°

x7 = 900°+180° = 1080°

We can see from the values above that the values of x that falls within the given range are 540°, 720° and 900°. This gives the required answers in degrees.

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.