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Question Number 12724 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 29/Apr/17

Answered by sma3l2996 last updated on 30/Apr/17

a^(cos^2 x) +a^(2cos^2 x−1) =a  a^(cos^2 x) +(a^(cos^2 x) )^2 ×a^(−1) =a  (a^(cos^2 x) )^2 +a×a^(cos^2 x) =a^2   (a^(cos^2 x) )^2 +2×(a/2)×a^(cos^2 x) +((a/2))^2 =((a/2))^2 +a^2   (a^(cos^2 x) +(a/2))^2 =(5/4)a^2   a^(cos^2 x) +(a/2)=+_− ((√5)/2)a  a^(cos^2 x) =−((1+(√5))/2)a   or  a^(cos^2 x) =((−1+(√5))/2)a  cos^2 x=log_a ((((√5)−1)/2)a)  x=acos((√(log_a ((((√5)−1)/2)a))))+2kπ   \k=(0,1,2,...)  for a=2  x=acos((√(log_2 ((√5)−1))))+2kπ  \k=(0,1,2,...)

$${a}^{{cos}^{\mathrm{2}} {x}} +{a}^{\mathrm{2}{cos}^{\mathrm{2}} {x}−\mathrm{1}} ={a} \\ $$$${a}^{{cos}^{\mathrm{2}} {x}} +\left({a}^{{cos}^{\mathrm{2}} {x}} \right)^{\mathrm{2}} ×{a}^{−\mathrm{1}} ={a} \\ $$$$\left({a}^{{cos}^{\mathrm{2}} {x}} \right)^{\mathrm{2}} +{a}×{a}^{{cos}^{\mathrm{2}} {x}} ={a}^{\mathrm{2}} \\ $$$$\left({a}^{{cos}^{\mathrm{2}} {x}} \right)^{\mathrm{2}} +\mathrm{2}×\frac{{a}}{\mathrm{2}}×{a}^{{cos}^{\mathrm{2}} {x}} +\left(\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} =\left(\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} +{a}^{\mathrm{2}} \\ $$$$\left({a}^{{cos}^{\mathrm{2}} {x}} +\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{5}}{\mathrm{4}}{a}^{\mathrm{2}} \\ $$$${a}^{{cos}^{\mathrm{2}} {x}} +\frac{{a}}{\mathrm{2}}=\underset{−} {+}\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}{a} \\ $$$${a}^{{cos}^{\mathrm{2}} {x}} =−\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}{a}\:\:\:{or}\:\:{a}^{{cos}^{\mathrm{2}} {x}} =\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}{a} \\ $$$${cos}^{\mathrm{2}} {x}={log}_{{a}} \left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}{a}\right) \\ $$$${x}={acos}\left(\sqrt{{log}_{{a}} \left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}{a}\right)}\right)+\mathrm{2}{k}\pi\:\:\:\backslash{k}=\left(\mathrm{0},\mathrm{1},\mathrm{2},...\right) \\ $$$${for}\:{a}=\mathrm{2} \\ $$$${x}={acos}\left(\sqrt{{log}_{\mathrm{2}} \left(\sqrt{\mathrm{5}}−\mathrm{1}\right)}\right)+\mathrm{2}{k}\pi\:\:\backslash{k}=\left(\mathrm{0},\mathrm{1},\mathrm{2},...\right) \\ $$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 30/Apr/17

thank you so much.

$${thank}\:{you}\:{so}\:{much}. \\ $$

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