Question-72248

Question Number 72248 by Omer Alattas last updated on 26/Oct/19

Commented by Omer Alattas last updated on 26/Oct/19

$${solv}\:{to}\:{x}\: \\ $$

Answered by MJS last updated on 26/Oct/19

tan^2 x −2sin 2x =0 x=arctan t t^2 −2((2t)/(t^2 +1))=0 ((t^2 (t^2 +1)−4t)/(t^2 +1))=0 t(t^3 +t−4)=0 t_1 =0 t_2 ≈1.37880 t_(3, 4) ≈−.689398±1.55750i x_1 =nπ x_2 ≈.943311+nπ

$$\mathrm{tan}^{\mathrm{2}} \:{x}\:−\mathrm{2sin}\:\mathrm{2}{x}\:=\mathrm{0} \\ $$$${x}=\mathrm{arctan}\:{t} \\ $$$${t}^{\mathrm{2}} −\mathrm{2}\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} +\mathrm{1}}=\mathrm{0} \\ $$$$\frac{{t}^{\mathrm{2}} \left({t}^{\mathrm{2}} +\mathrm{1}\right)−\mathrm{4}{t}}{{t}^{\mathrm{2}} +\mathrm{1}}=\mathrm{0} \\ $$$${t}\left({t}^{\mathrm{3}} +{t}−\mathrm{4}\right)=\mathrm{0} \\ $$$${t}_{\mathrm{1}} =\mathrm{0} \\ $$$${t}_{\mathrm{2}} \approx\mathrm{1}.\mathrm{37880} \\ $$$${t}_{\mathrm{3},\:\mathrm{4}} \approx−.\mathrm{689398}\pm\mathrm{1}.\mathrm{55750i} \\ $$$${x}_{\mathrm{1}} ={n}\pi \\ $$$${x}_{\mathrm{2}} \approx.\mathrm{943311}+{n}\pi \\ $$

Commented by Omer Alattas last updated on 27/Oct/19

$${thank}\:{you}\:{pro} \\ $$

Commented by Omer Alattas last updated on 26/Oct/19

$${how}\:{i}\:{can}\:{write}\:{the}\:{anccer}\:{in}\:{rad}\:{pleas} \\ $$

Commented by MJS last updated on 27/Oct/19

x∈[0; 2π] x_1 leads to 0; π; 2π x_2 leads to .943311; 4.08490

$${x}\in\left[\mathrm{0};\:\mathrm{2}\pi\right] \\ $$$${x}_{\mathrm{1}} \:\mathrm{leads}\:\mathrm{to}\:\mathrm{0};\:\pi;\:\mathrm{2}\pi \\ $$$${x}_{\mathrm{2}} \:\mathrm{leads}\:\mathrm{to}\:.\mathrm{943311};\:\mathrm{4}.\mathrm{08490} \\ $$

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