Question

How do you find all `(cos2x)/(sin3x-sinx)` in the interval `[0,2pi)`?

Answer 1

Graph reads y values, `x in [ 0, 2pi ]`
sans indeterminate holes at `x = pi/4, 3/4pi, 5/4pi and 7/4pi`, and
asymptotic `x = 0, pi and 2pi`

Explanation:

`y = (cos 2x)/(sin 3x - sin x)`

`= (cos 2x)/(2 cos(1/2(3x + x ))sin(1/2( 3x - x )))`

`=1/2 (cos 2x)/(cos 2x)(1/sinx),`

`= 1/2 csc x, cos 2x ne 0`
`rArr 2x ne (2k + 1 ) (pi/2), k = 0, +-1, +-2, +-3, ...`

`x ne (2k + 1 )pi/4 and` asymptotic `x ne kpi`

Note that

`y = 1/2 csc x notin 1/2 ( - 1, 1 ) = ( - 1/2, 1/2 )`

The graph reads y. sans y at duly marked `x =0, pi/4, 3/4pi,`

`pi, 5/4pi, 7/4pi, 2pi`. .
graph{(2y sin x +1)(x-pi/4+0.001y)(x+0.001y)(x-pi+0.001y) (x-3pi/4+0.001y)(x-5pi/4+0.001y)(x-2pi+0.001y)(x-7pi/4+0.001y)= 0[-0.2 8 -2 2]}