Question

Rezolvați ecuația :

$sin4x = cos6x$

Answer 1

Pentru inceput facem substiutia: u=2x.Atunci :

$\sin 2u=\cos 3u$

Aplicam formulele:

$\sin 2u=2*\sin u*\cos u\\\cos 3u= \cos^3 u-3*\sin^2 u*\cos u$

Inlocuim in ecuatia data:

$2*\sin u*\cos u-\cos^3 u+3*\sin^2 u*\cos u=0\\\cos u*(3*\sin^2 u+2*\sin u-\cos^2 u)=0$

Distingem doua cazuri posibile :

Cazul 1:

$\cos u=0\Rightarrow u=\pm \frac{\pi}{2}+2k\pi ,k\in Z$

Cazul 2:

$3*\sin^2 u+2*\sin u-\cos^2 u=0\\3*\sin^2 u+2*\sin u-1+\sin^2 u=0\\4*\sin^2 u+2*\sin u-1=0\\\sin u=t (not)\\4t^2+2t-1=0\\\Delta=4+16=20\Rightarrow \sqrt{\Delta}=2\sqrt 5\\t_1= \frac{-2+2\sqrt 5}{8}=\frac{\sqrt 5-1}{4}\\t_2=\frac{-2-2\sqrt 5}{8}= - \frac{\sqrt 5+1}{4}\\Revenim~ la~ substitutie~ :\\\sin u=\frac{\sqrt 5-1}{4}\Rightarrow u=(-1)^k*\arcsin (\frac{\sqrt 5-1}{4})+k*\pi ,k\in Z\\ \sin u=-\frac{\sqrt 5+1}{4}\Rightarrow u=(-1)^{k+1}*\arcsin (\frac{\sqrt 5+1}{2} )+k*\pi,k\in Z$

Obtinem solutia:

$u\in \{\pm \frac{\pi}{2}+2k\pi\}\cup \{(-1)^k*\arcsin (\frac{\sqrt 5-1}{4})+k*\pi\}\cup\\ \{(-1)^{k+1} *\arcsin(\frac{\sqrt 5+1}{4})+k*\pi\},k\in Z\\Revenim~ la~ substitutia~ principala~ si~ obtinem : \\x\in \{\pm \frac{\pi}{4}+k\pi\}\cup \{\frac{(-1)^k*\arcsin (\frac{\sqrt 5-1}{4})+k*\pi}{2}\}\cup\{\frac{(-1)^{k+1} *\arcsin(\frac{\sqrt 5+1}{4})+k*\pi}{2}\},\\k\in Z$