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

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..." was answered by our experts

idiopatia0f

Answered question

2022-01-14

Equation

x = \sqrt{2 + \sqrt{2 - \sqrt{2 + x}}}

Using

x = 2 \cos (θ)

we get

2 \cos (θ) = \sqrt{2 + \sqrt{2 - \sqrt{2 + 2 \cos θ}}}

2 \cos θ = \sqrt{2 + \sqrt{2 - 2 \cos \frac{θ}{2}}}

2 \cos θ = \sqrt{2 + 2 \sin \frac{θ}{4}}

4 \cos^{2} θ = 2 + 2 \sin \frac{θ}{4}

After this step I am not able to solve it

Answer & Explanation

David Clayton

Beginner2022-01-15Added 36 answers

My illustration is outlined below

4 \cos^{2} θ = 2 (1 + \sin \frac{θ}{4})

4 \cos^{2} 10^{\circ} \neq 2 (1 + \sin \frac{10^{\circ}}{4})

4 \cos^{2} 10^{\circ} \neq 2 (1 + {\sin 2.5}^{\circ})

4 \cos^{2} 10^{\circ} \neq 2 (1 + {\cos 87.5}^{\circ})

4 \cos^{2} 20^{\circ} \neq 2 (1 + \sin \frac{10^{\circ}}{4})

4 \cos^{2} 20^{\circ} \neq 2 (1 + {\sin 5}^{\circ})

4 \cos^{2} 20^{\circ} \neq 2 (1 + {\cos 85}^{\circ})

4 \cos^{2} 40^{\circ} \neq 2 (1 + \sin \frac{40^{\circ}}{4})

4 \cos^{2} 40^{\circ} \neq 2 (1 + {\sin 10}^{\circ})

4 \cos^{2} 40^{\circ} \neq 2 (1 + {\cos 80}^{\circ})

4 \cos^{2} 80^{\circ} \neq 2 (1 + \sin \frac{80^{\circ}}{4})

4 \cos^{2} 80^{\circ} \neq 2 (1 + {\sin 20}^{\circ})

4 \cos^{2} 80^{\circ} \neq 2 (1 + {\cos 70}^{\circ})

Hence the option is

0 = 40^{\circ}

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