Question

Cum se rezolvă?
$3 \cos(3x) + 5 \sin(3x) = 5$

Answer 1

Răspuns:

Explicație pas cu pas:

$cos\alpha =\frac{1-tg^{2}\frac{\alpha}{2} }{1+tg^{2}\frac{\alpha}{2} },~sin\alpha =\frac{2tg\frac{\alpha}{2} }{1+tg^{2}\frac{\alpha}{2} } \\3*\frac{1-tg^{2}\frac{3\alpha}{2} }{1+tg^{2}\frac{3\alpha}{2} }+5*\frac{2tg\frac{3\alpha}{2} }{1+tg^{2}\frac{3\alpha}{2} } =5~|*(1+tg^{2}\frac{3\alpha}{2}) \\3*(1-tg^{2}\frac{3\alpha}{2} )+10*tg\frac{3\alpha}{2} =5*(1+tg^{2}\frac{3\alpha}{2})\\fie~tg\frac{3\alpha}{2}=t,~atunci~avem\\3*(1-t^{2})+10t=5(1+t^{2}.~3-3t^{2}+10t=5+5t^{2},~8t^{2}-10t+2=0,~:2$

$4t^{2}-5t+1=0,~delta=9,~t=\frac{1}{4} ~sau~t=1~atunci~avem:\\tg\frac{3\alpha }{2}=\frac{1}{4},~ \frac{3\alpha }{2}=arctg\frac{1}{4} +k\pi ,~\alpha =\frac{2}{3}*arctg\frac{1}{4} + \frac{2\pi }{3}*k,~k~apartine~Z\\ sau ~tg\frac{3\alpha }{2}=1,~\frac{3\alpha }{2}}=arctg1+\pi k,~\frac{3\alpha }{2}=\frac{\pi}{4}+\pi*k,~\alpha =\frac{\pi }{6} +\frac{2\pi }{3}*k$