Solve the equation by factoring. (enter your answers as a comma-separated list. let k be any integer. round terms to three decimal places where appropriate. if there is no solution, enter no solution.) csc(θ) cot(θ) − sin(θ) tan(θ) = cos(θ)

There is no solution

Step-by-step explanation:

Given that

csc(θ) cot(θ) − sin(θ) tan(θ) = cos(θ) (1)

Where

csc(θ) = 1/sin(θ)

cot(θ) = 1/tan(θ) = cos(θ)/sin(θ)

tan(θ) = sin(θ)/cos(θ)

Substitute them in equation (1)

1/sin(θ) × cos(θ)/sin(θ) - sin(θ)×sin(θ)/cos(θ) = cos(θ)

Cos(θ)/sin^2(θ) - sin^2(θ)/cos(θ) = cos(θ)

Cos(θ)/sin^2(θ) = sin^2(θ)/cos(θ) + cos(θ)

Take the LCM of right hand side

Cos(θ)/sin^2(θ) = (sin^2(θ) + cos^2(θ))/ cos(θ)

But sin^2(θ) + cos^2(θ) = 1

Cos(θ)/sin^2(θ) = 1 /cos(θ)

Cross multiply

Cos^2(θ) = sin^2(θ)

Cos^2(θ) - sin^2(θ) = 0

If θ = 45

That means there's no solution

b. x- 48 =120

step-by-step explanation:

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'.