1.) 3 sin²x = sin2x cosx
3 ( 1-cos²x) = 2sinx cosx cosx
2.) tgx -1/tgx - 2/√3 = 0
y - 1/y - 2/√3 = 0
Hi, I try to solve these equations but Idon't know what I should do next. Can you help me? Thanks(Wink)
1.
$\displaystyle \displaystyle \begin{align*} 3\sin^2{(x)} &= \sin{(2x)}\cos{(x)} \\ 3\sin^2{(x)} &= 2\sin{(x)}\cos{(x)}\cos{(x)} \\ 3\sin^2{(x)} &= 2\sin{(x)}\cos^2{(x)} \\ 3\sin^2{(x)} &= 2\sin{(x)} \left[ 1 - \sin^2{(x)} \right] \\ 3\sin^2{(x)} &= 2\sin{(x)} - 2\sin^3{(x)} \\ 2\sin^3{(x)} + 3\sin^2{(x)} - 2\sin{(x)} &= 0 \\ \end{align*}$
This can now be solved by factorising.
1.) I want to use a substitution but I can't because of 2 sinx. I need cosx there. cosx² I can write like 1-sinx² but there is only cosx. I must chang and I must change cosx to sinx but I don't know how.
2.) y² - 1 - 2/√3y = 0
y² - 2/√3y - 1 = 0
D= b²- 4ac
D= (-2/√3)²-4 * 1 * (-1)
√D = 4 √3/3
x1,2 = -b ± √D/2a
x1,2 = 2/ √3 ± 4 √3/3:2
x1 = √3....................tg is π/3
x2= - √3/3..............tg is 5/6π and 11/6π. Is that solition right?