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Question Number 69535 by naka3546 last updated on 24/Sep/19

Commented by naka3546 last updated on 24/Sep/19

$w i t h o u t u$
	Answered by MJS last updated on 24/Sep/19

	$a \tan t = k (1 + \sec t)$ $\Rightarrow$ $k + k \cos t - a \sin t = 0$ $t = 2 \arctan u$ $2 \frac{k - a u}{u^{2} + 1} = 0 \Rightarrow u = \frac{k}{a} \Rightarrow t = 2 \arctan \frac{k}{a}$ $\Rightarrow$ $2 \arctan \frac{k}{3} + 2 \arctan \frac{k}{4} + 2 \arctan \frac{k}{5} + 2 \arctan \frac{k}{6} = 2 π$ $\arctan \frac{k}{3} + \arctan \frac{k}{4} + \arctan \frac{k}{5} + \arctan \frac{k}{6} = π$ $\tan (\arctan a + \arctan b) = \frac{a + b}{1 - a b}$ $\tan (\arctan a + \arctan b + \arctan c) =$ $= \frac{a + b + c - a b c}{1 - (a b + a c + b c)}$ $\tan (\arctan a + \arctan b + \arctan c + \arctan d) =$ $= \frac{a + b + c + d - (a b c + a b d + a c d + b c d)}{1 + a b c d - (a b + a c + a d + b c + b d d + c d)}$ $\Rightarrow$ $- \frac{18 k (k^{2} - 19)}{k^{4} - 119 k^{2} + 360} = 0 \Rightarrow k = \pm \sqrt{19} \lor k = 0$ $but k = 0 \Rightarrow w = x = y = z = 0$ $k = - \sqrt{19} {doesn}^{'} t fit the above equation$ $\Rightarrow$ $k = \sqrt{19}$
	Commented by mind is power last updated on 24/Sep/19

	$n i c e s o l u t i o n s i r, k = 0 w e c a n h a v e π, π, 0, 0$
	Commented by MJS last updated on 24/Sep/19

	$true$
	Answered by mr W last updated on 24/Sep/19
	$sin w=(k/3)(1+cos w) sin^2 w=(k^2 /3^2 )(1+cos^2 w+2 cos w) ⇒(1+(3^2 /k^2 ))cos^2 w+2 cos w+(1−(3^2 /k^2 ))=0 ⇒cos w=((−1±(3^2 /k^2 ))/(1+(3^2 /k^2 )))= { ((−1 ⇒w=π)),(((3^2 −k^2 )/(3^2 +k^2 ))) :} similarly ⇒cos x= { ((−1 ⇒x=π)),(((4^2 −k^2 )/(4^2 +k^2 ))) :} ⇒cos y= { ((−1 ⇒y=π)),(((5^2 −k^2 )/(5^2 +k^2 ))) :} ⇒cos z= { ((−1 ⇒z=π)),(((6^2 −k^2 )/(6^2 +k^2 ))) :} .... k=±(√(19))=±4.3589 ???$
	$\sin w = \frac{k}{3} (1 + \cos w)$ $\sin^{2} w = \frac{k^{2}}{3^{2}} (1 + \cos^{2} w + 2 \cos w)$ $\Rightarrow (1 + \frac{3^{2}}{k^{2}}) \cos^{2} w + 2 \cos w + (1 - \frac{3^{2}}{k^{2}}) = 0$ $\Rightarrow \cos w = \frac{- 1 \pm \frac{3^{2}}{k^{2}}}{1 + \frac{3^{2}}{k^{2}}} = {\begin{cases} - 1 \Rightarrow w = π \\ \frac{3^{2} - k^{2}}{3^{2} + k^{2}} \end{cases}$ $s i m i l a r l y$ $\Rightarrow \cos x = {\begin{cases} - 1 \Rightarrow x = π \\ \frac{4^{2} - k^{2}}{4^{2} + k^{2}} \end{cases}$ $\Rightarrow \cos y = {\begin{cases} - 1 \Rightarrow y = π \\ \frac{5^{2} - k^{2}}{5^{2} + k^{2}} \end{cases}$ $\Rightarrow \cos z = {\begin{cases} - 1 \Rightarrow z = π \\ \frac{6^{2} - k^{2}}{6^{2} + k^{2}} \end{cases}$ $. . . .$ $k = \pm \sqrt{19} = \pm 4.3589 ? ? ?$
	Commented by mind is power last updated on 24/Sep/19

	$n i c e s i r p l e a s e h o w h o w y o u p u t {f o r t w o l i g n s / ?$
	Commented by mr W last updated on 24/Sep/19

	$j u s t u s e t h e t h i r d b u t t o n f r o m l e f t o n$ $t h e b o t t o m o f m e n u .$ $e x a m p l e {\begin{cases} l i n e 1 \\ l i n e 2 \end{cases}$
	Commented by mind is power last updated on 24/Sep/19

	$t h a n x$
	Answered by mind is power last updated on 24/Sep/19
	$⇒3sin(w)=k(cos(w)+1) 4sin(x)=k(cos(x)+1) 5sin(y)=k(cos(y)+1) 6sin(z)=k(cos(z)+1) ⇔3tg((w/2))=4tg((x/2))=5tg((y/2))=6tg((z/2))=k tg(a)+tg(b)=((sin(a+b))/(cos(a)cos(b))) ⇒tg(((x+w)/2))+tg(((z+y)/2))=((sin(((x+y+z+w)/2)))/(cos(((x+y)/2))cos(((z+w)/2)))) tg(a+b)=((tg(a)+tg(b))/(1−tg(a)tg(b))) tg(((x+w)/2))+tg(((y+z)/2))=((tg((x/2))+tg((w/2)))/(1−tg((x/2))tg((w/2))))+((tg((y/2))+tg((z/2)))/(1−tg((y/2)).tg((z/2))))=0 we have tg((w/2))=(k/3) tg((x/2))=(k/4) , tg((y/2))=(k/5), tg((z/2))=(k/6) ⇒(((k/3)+(k/4))/(1−(k^2 /(12))))+(((k/5)+(k/6))/(1−(k^2 /(30))))=0 ⇒((7k)/(12−k^2 ))+((11k)/(30−k^2 ))=0 ⇒k(((7.(30−k^2 )+11(12−k^2 ))/((12−k^2 )(30−k^2 ))))=0 ⇒k=0 or 210−7k^2 +132−11k^2 =0⇒k^2 =((342)/(18))=19 ⇒k=+_− (√(19)) k=−(√(19))⇒(w/2) ,(x/2),(y/2),(x/2)>(π/2)⇒x+y+z+w>4π>2π k=(√(19)) i will study this later k∈{0,(√(19))}$
	$\Rightarrow 3 s i n (w) = k (c o s (w) + 1)$ $4 s i n (x) = k (c o s (x) + 1)$ $5 s i n (y) = k (c o s (y) + 1)$ $6 s i n (z) = k (c o s (z) + 1)$ $\Leftrightarrow 3 t g (\frac{w}{2}) = 4 t g (\frac{x}{2}) = 5 t g (\frac{y}{2}) = 6 t g (\frac{z}{2}) = k$ $t g (a) + t g (b) = \frac{s i n (a + b)}{c o s (a) c o s (b)}$ $\Rightarrow t g (\frac{x + w}{2}) + t g (\frac{z + y}{2}) = \frac{s i n (\frac{x + y + z + w}{2})}{c o s (\frac{x + y}{2}) c o s (\frac{z + w}{2})}$ $t g (a + b) = \frac{t g (a) + t g (b)}{1 - t g (a) t g (b)}$ $t g (\frac{x + w}{2}) + t g (\frac{y + z}{2}) = \frac{t g (\frac{x}{2}) + t g (\frac{w}{2})}{1 - t g (\frac{x}{2}) t g (\frac{w}{2})} + \frac{t g (\frac{y}{2}) + t g (\frac{z}{2})}{1 - t g (\frac{y}{2}) . t g (\frac{z}{2})} = 0$ $w e h a v e t g (\frac{w}{2}) = \frac{k}{3}$ $t g (\frac{x}{2}) = \frac{k}{4}, t g (\frac{y}{2}) = \frac{k}{5}, t g (\frac{z}{2}) = \frac{k}{6}$ $\Rightarrow \frac{\frac{k}{3} + \frac{k}{4}}{1 - \frac{k^{2}}{12}} + \frac{\frac{k}{5} + \frac{k}{6}}{1 - \frac{k^{2}}{30}} = 0$ $\Rightarrow \frac{7 k}{12 - k^{2}} + \frac{11 k}{30 - k^{2}} = 0$ $\Rightarrow k (\frac{7 . (30 - k^{2}) + 11 (12 - k^{2})}{(12 - k^{2}) (30 - k^{2})}) = 0$ $\Rightarrow k = 0 o r 210 - 7 k^{2} + 132 - 11 k^{2} = 0 \Rightarrow k^{2} = \frac{342}{18} = 19$ $\Rightarrow k = \underline{+} \sqrt{19}$ $k = - \sqrt{19} \Rightarrow \frac{w}{2}, \frac{x}{2}, \frac{y}{2}, \frac{x}{2} > \frac{π}{2} \Rightarrow x + y + z + w > 4 π > 2 π$ $k = \sqrt{19} i w i l l s t u d y t h i s l a t e r k \in {0, \sqrt{19}}$
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