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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
	$a^(cos^2 x) +a^(2cos^2 x−1) =a a^(cos^2 x) +(a^(cos^2 x) )^2 ×a^(−1) =a (a^(cos^2 x) )^2 +a×a^(cos^2 x) =a^2 (a^(cos^2 x) )^2 +2×(a/2)×a^(cos^2 x) +((a/2))^2 =((a/2))^2 +a^2 (a^(cos^2 x) +(a/2))^2 =(5/4)a^2 a^(cos^2 x) +(a/2)=+_− ((√5)/2)a a^(cos^2 x) =−((1+(√5))/2)a or a^(cos^2 x) =((−1+(√5))/2)a cos^2 x=log_a ((((√5)−1)/2)a) x=acos((√(log_a ((((√5)−1)/2)a))))+2kπ \k=(0,1,2,...) for a=2 x=acos((√(log_2 ((√5)−1))))+2kπ \k=(0,1,2,...)$
	$a^{{c o s}^{2} x} + a^{2 {c o s}^{2} x - 1} = a$ $a^{{c o s}^{2} x} + {(a^{{c o s}^{2} x})}^{2} \times a^{- 1} = a$ ${(a^{{c o s}^{2} x})}^{2} + a \times a^{{c o s}^{2} x} = a^{2}$ ${(a^{{c o s}^{2} x})}^{2} + 2 \times \frac{a}{2} \times a^{{c o s}^{2} x} + {(\frac{a}{2})}^{2} = {(\frac{a}{2})}^{2} + a^{2}$ ${(a^{{c o s}^{2} x} + \frac{a}{2})}^{2} = \frac{5}{4} a^{2}$ $a^{{c o s}^{2} x} + \frac{a}{2} = \underline{+} \frac{\sqrt{5}}{2} a$ $a^{{c o s}^{2} x} = - \frac{1 + \sqrt{5}}{2} a o r a^{{c o s}^{2} x} = \frac{- 1 + \sqrt{5}}{2} a$ ${c o s}^{2} x = {l o g}_{a} (\frac{\sqrt{5} - 1}{2} a)$ $x = a c o s (\sqrt{{l o g}_{a} (\frac{\sqrt{5} - 1}{2} a)}) + 2 k π ∖ k = (0, 1, 2, . . .)$ $f o r a = 2$ $x = a c o s (\sqrt{{l o g}_{2} (\sqrt{5} - 1)}) + 2 k π ∖ k = (0, 1, 2, . . .)$
	Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 30/Apr/17

	$t h a n k y o u s o m u c h .$
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