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)

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- Feb 22nd 2013, 05:12 AMAtalantetrigonometric equations
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) - Feb 22nd 2013, 06:03 AMHallsofIvyRe: trigonometric equations
It would help if you told us

**what**you tried and where you got stuck.

(Substituting y for tan(x) in (2) was a good idea. Now multiply both sides of the equation by y.) - Feb 22nd 2013, 07:01 AMProve ItRe: trigonometric equations
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. - Feb 22nd 2013, 07:18 AMAtalanteRe: trigonometric equations
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? - Feb 22nd 2013, 06:01 PMProve ItRe: trigonometric equations
- Feb 23rd 2013, 03:16 AMAtalanteRe: trigonometric equations
It's ok now:-) The post came while I was writing. Thank you very much.