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Question-75901




Question Number 75901 by Master last updated on 20/Dec/19
Answered by MJS last updated on 20/Dec/19
(1/(cos^2  x))=4tan 2x  (1/(1+cos 2x))=2tan 2x  x=arctan t  ((t^2 +1)/2)=((4t)/(1−t^2 ))  t^4 +8t−1=0  t_1 ≈−2.04004  t_2 ≈.124970  t_(3, 4) ≈.957536±1.73366i  ⇒  x=nπ−1.11503∨x=nπ+.124325 with n∈Z
$$\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:{x}}=\mathrm{4tan}\:\mathrm{2}{x} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+\mathrm{cos}\:\mathrm{2}{x}}=\mathrm{2tan}\:\mathrm{2}{x} \\ $$$${x}=\mathrm{arctan}\:{t} \\ $$$$\frac{{t}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{4}{t}}{\mathrm{1}−{t}^{\mathrm{2}} } \\ $$$${t}^{\mathrm{4}} +\mathrm{8}{t}−\mathrm{1}=\mathrm{0} \\ $$$${t}_{\mathrm{1}} \approx−\mathrm{2}.\mathrm{04004} \\ $$$${t}_{\mathrm{2}} \approx.\mathrm{124970} \\ $$$${t}_{\mathrm{3},\:\mathrm{4}} \approx.\mathrm{957536}\pm\mathrm{1}.\mathrm{73366i} \\ $$$$\Rightarrow \\ $$$${x}={n}\pi−\mathrm{1}.\mathrm{11503}\vee{x}={n}\pi+.\mathrm{124325}\:\mathrm{with}\:{n}\in\mathbb{Z} \\ $$
Commented by Master last updated on 21/Dec/19
Commented by MJS last updated on 21/Dec/19
believe me or not  try the given options, they are all wrong
$$\mathrm{believe}\:\mathrm{me}\:\mathrm{or}\:\mathrm{not} \\ $$$$\mathrm{try}\:\mathrm{the}\:\mathrm{given}\:\mathrm{options},\:\mathrm{they}\:\mathrm{are}\:\mathrm{all}\:\mathrm{wrong} \\ $$
Commented by MJS last updated on 21/Dec/19
I guess the question is wrong  (1/(cos^2  x))=4tan^2  x  (1/(cos^2  x))=4((sin^2  x)/(cos^2  x))  cos^2  x ≠0 ⇒ x≠(π/2)+nπ  sin^2  x =(1/4)  sin x =±(1/2)  x=±(π/6)+nπ
$$\mathrm{I}\:\mathrm{guess}\:\mathrm{the}\:\mathrm{question}\:\mathrm{is}\:\mathrm{wrong} \\ $$$$\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:{x}}=\mathrm{4tan}^{\mathrm{2}} \:{x} \\ $$$$\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:{x}}=\mathrm{4}\frac{\mathrm{sin}^{\mathrm{2}} \:{x}}{\mathrm{cos}^{\mathrm{2}} \:{x}} \\ $$$$\mathrm{cos}^{\mathrm{2}} \:{x}\:\neq\mathrm{0}\:\Rightarrow\:{x}\neq\frac{\pi}{\mathrm{2}}+{n}\pi \\ $$$$\mathrm{sin}^{\mathrm{2}} \:{x}\:=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{sin}\:{x}\:=\pm\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${x}=\pm\frac{\pi}{\mathrm{6}}+{n}\pi \\ $$
Commented by Master last updated on 21/Dec/19
thanks
$$\mathrm{thanks} \\ $$

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