Math Question and Answers


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 190938 by Shrinava last updated on 14/Apr/23

	Answered by Frix last updated on 15/Apr/23
	$tan 3x −tan x tan 11x tan 13x =0 We can transform this to sin 28x −sin 20x +sin 4x =0 y=4x sin 7y −sin 5y +sin y =0 s=sin y & transforming s(s^6 −((3s^4 )/2)+((9s^2 )/(16))−(3/(64)))=0 s=0 ★ t=s^2 t^3 −((3t^2 )/2)+((9t)/(16))−(3/(64))=0 u=t−(1/2) u^3 −((3u)/(16))−(1/(64))=0 u=−((cos ((2π)/9))/2)∨u=−((sin (π/(18)))/2)∨u=((cos (π/9))/2) ★ Now we have to go back For 0≤x<(π/2) I get (x/π)∈{0, (1/(36)), (1/(18)), (1/9), (5/(36)), (7/(36)), (2/9), (1/4), (5/(18)), ((11)/(36)), ((13)/(36)), (7/(18)), (4/9), ((17)/(36))}$
	$\tan 3 x - \tan x \tan 11 x \tan 13 x = 0$ $We can transform this to$ $\sin 28 x - \sin 20 x + \sin 4 x = 0$ $y = 4 x$ $\sin 7 y - \sin 5 y + \sin y = 0$ $s = \sin y & transforming$ $s (s^{6} - \frac{3 s^{4}}{2} + \frac{9 s^{2}}{16} - \frac{3}{64}) = 0$ $s = 0 ★$ $t = s^{2}$ $t^{3} - \frac{3 t^{2}}{2} + \frac{9 t}{16} - \frac{3}{64} = 0$ $u = t - \frac{1}{2}$ $u^{3} - \frac{3 u}{16} - \frac{1}{64} = 0$ $u = - \frac{\cos \frac{2 π}{9}}{2} \lor u = - \frac{\sin \frac{π}{18}}{2} \lor u = \frac{\cos \frac{π}{9}}{2} ★$ $Now we have to go back$ $For 0 ⩽ x < \frac{π}{2} I get$ $\frac{x}{π} \in {0, \frac{1}{36}, \frac{1}{18}, \frac{1}{9}, \frac{5}{36}, \frac{7}{36}, \frac{2}{9}, \frac{1}{4}, \frac{5}{18}, \frac{11}{36}, \frac{13}{36}, \frac{7}{18}, \frac{4}{9}, \frac{17}{36}}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum