Question

2sinx+cotx-cscx=0

Answer 1

The equation can be written in the form:
2sin x + cos x/sin x - 1/sin x = 0. Make sin x common denominator
2sin^2 x + cos x - 1 = 0 -> 2(1 - cos^2 x) + cos x - 1 = 0
Call cos x = x. We get a quadratic equation: -2x*2 + x + 1 = 0 (1)
2 Tips to remember when quick solving quadratic equations:
Tip 1. When a + b + c = 0, one real root is (1) and the other is (c/a)
Tip 2. When a - b + c = 0, one real root is (-1), and the other is (-c/a)
The equation (1) has a + b + c = 0, then one real root is (1) and the other is c/a = -1/2.
Next, solve for x.
x1 = cos x = 1 = cos 0 and cos 2Pi -> x1 = 0 and ....
x2 = cos x =-1/2 = cos (2Pi/3) and cos (4Pi/3)-> x2 = 2Pi/3 and .....