Equation x=2+2−2+x Using x=2cos⁡(θ) we get 2cos⁡(θ)=2+2−2+2cos⁡θ 2cos⁡θ=2+2−2cos⁡θ2 2cos⁡θ=2+2sin⁡θ4 4cos2θ=2+2sin⁡θ4 After this step I am not able to solve it

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Equation x=2+2−2+x  Using x=2cos⁡(θ) we get 2cos⁡(θ)=2+2−2+2cos⁡θ  2cos⁡θ=2+2−2cos⁡θ2   2cos⁡θ=2+2sin⁡θ4   4cos2θ=2+2sin⁡θ4  After this step I am not able to solve it

David Clayton · Accepted Answer

My illustration is outlined below  4cos2θ=2(1+sin⁡θ4)   4cos210∘≠2(1+sin⁡10∘4)    4cos210∘≠2(1+sin⁡2.5∘)    4cos210∘≠2(1+cos⁡87.5∘)    4cos220∘≠2(1+sin⁡10∘4)    4cos220∘≠2(1+sin⁡5∘)    4cos220∘≠2(1+cos⁡85∘)    4cos240∘≠2(1+sin⁡40∘4)    4cos240∘≠2(1+sin⁡10∘)    4cos240∘≠2(1+cos⁡80∘)    4cos280∘≠2(1+sin⁡80∘4)    4cos280∘≠2(1+sin⁡20∘)    4cos280∘≠2(1+cos⁡70∘)    Hence the option is  0=40∘

Equation x=2+2−2+x Using x=2cos⁡(θ) we get 2cos⁡(θ)=2+2−2+2cos⁡θ 2cos⁡θ=2+2−2cos⁡θ2 2cos⁡θ=2+2sin⁡θ4 4cos2θ=2+2sin⁡θ4 After this step I am...

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