Solve the equation: (log _(sin x) cos x)^2 =1


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Question Number 20102 by ajfour last updated on 21/Aug/17

$S o l v e t h e e q u a t i o n :$ ${(\log_{\sin x} \cos x)}^{2} = 1$
	Answered by allizzwell23 last updated on 21/Aug/17

	$\log_{\sin x} \cos x = 1$ $\cos x = (\sin x)$ $\cos x = \sin x$ $x = \frac{π}{4}$
	Answered by sma3l2996 last updated on 21/Aug/17
	$(((ln(cosx))/(ln(sinx))))^2 =1⇔(ln(cosx))^2 =(ln(sinx))^2 ln(cosx)=ln(sinx) or ln(cosx)=−ln(sinx) so x=(π/4)+2kπ or cosx=(1/(sinx)) and that imposible so than x=π/4+2kπ \ k=(0,1,2,3,...)$
	${(\frac{l n (c o s x)}{l n (s i n x)})}^{2} = 1 \Leftrightarrow {(l n (c o s x))}^{2} = {(l n (s i n x))}^{2}$ $l n (c o s x) = l n (s i n x) o r l n (c o s x) = - l n (s i n x)$ $s o x = \frac{π}{4} + 2 k π o r c o s x = \frac{1}{s i n x} a n d t h a t i m p o s i b l e$ $s o t h a n x = π / 4 + 2 k π ∖ k = (0, 1, 2, 3, . . .)$
	Commented by sma3l2996 last updated on 21/Aug/17

	$y o u w e l c o m e$
	Commented by ajfour last updated on 21/Aug/17

	$t h a n k s s i r, f o r g e n e r a l s o l u t i o n .$
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