How many solutions to z6!−z5!∈R and |z|=1The following is my interpretation of the first solution.The next step is to see how many solutions to sin⁡(720θ)−sin⁡(120θ)=0. (1)One way to do this is consider sin⁡(6ω)−sin⁡(ω), which has 12 solutions. (2)For each solution to (2), corresponds 120 solutions to (1). I understand that.But how do I know that sin⁡(6ω)=sin⁡(ω) has 12 solutions?

Question

How many solutions to z6!−z5!∈R  and  |z|=1The following is my interpretation of the first solution.The next step is to see how many solutions to sin⁡(720θ)−sin⁡(120θ)=0. (1)One way to do this is consider sin⁡(6ω)−sin⁡(ω), which has 12 solutions. (2)For each solution to (2), corresponds 120 solutions to (1). I understand that.But how do I know that sin⁡(6ω)=sin⁡(ω) has 12 solutions?

tigging9k0 · Accepted Answer

sin⁡6w=sin⁡w⇔6w=2kπ+w  or  6w=2kπ+π−w⇔w=2kπ5  or  w=2k+17πwith k∈Z, thenw=0,2π5,4π5,6π5,8π5,π7,3π7,5π7,7π7,9π7,11π7,13π7

elseptimopc7 · Accepted Answer

Sincesin⁡(θ)=sin⁡(π−θ)=sin⁡(3π−θ)=sin⁡(5π−θ)=sin⁡(7π−θ)=sin⁡(9π−θ)=sin⁡(11π−θ)=sin⁡(13π−θ⏟6θ)if 6θ equals any of the arguments to sin⁡x on the right, then sin⁡(θ)=sin⁡(6θ). That is,θ∈{π7,3π7,5π7,π,9π7,11π7,13π7}

How many solutions to z^{6!}-z^{5!} \in \mathbb{R} \text{ and } |z| =1 The following is my

Answered question

Answer & Explanation

New Questions in Precalculus