Solving periodic equation cos⁡7θ=cos⁡3θ+sin⁡5θI approached the problem as follows : cos⁡7θ=cos⁡3θ+sin⁡5θ⇒cos⁡7θ−cos⁡3θ=sin⁡5θ⇒2sin⁡5θsin⁡(−2θ)=sin⁡5θFrom there on we get, sin⁡2θ=−12 and sin⁡5θ=0So, we deduce that θ=−π12+πk and θ=25πn where k and n are integers.However, Wolfram|Alpha doesn't seem to agree with me. They present π5 as a solution which is not included in my solution.

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Solving periodic equation cos⁡7θ=cos⁡3θ+sin⁡5θI approached the problem as follows : cos⁡7θ=cos⁡3θ+sin⁡5θ⇒cos⁡7θ−cos⁡3θ=sin⁡5θ⇒2sin⁡5θsin⁡(−2θ)=sin⁡5θFrom there on we get, sin⁡2θ=−12  and  sin⁡5θ=0So, we deduce that θ=−π12+πk and θ=25πn where k and n are integers.However, Wolfram|Alpha doesn&#039;t seem to agree with me. They present π5 as a solution which is not included in my solution.

Alejandra Hanna · Accepted Answer

sin⁡5θ=0sin⁡x=0 has solutions x=nπThus we get5θ=nπθ=nπ5   n∈ZAlsosin⁡2θ=−122θ=2nπ−π6⇒θ=nπ−π12,    n∈Z2θ=(2n+1)π+π6⇒θ=nπ+7π12,    n∈Z

Solving periodic equation \(\displaystyle{\cos{{7}}}\theta={\cos{{3}}}\theta+{\sin{{5}}}\theta\) I approached the problem

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