# Math Help - trigonometric equations

1. ## trigonometric 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)

2. ## Re: 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.)

3. ## Re: trigonometric equations

Originally Posted by Atalante
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
1.
\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.

4. ## Re: 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?

5. ## Re: trigonometric equations

Originally Posted by Atalante
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.
Did you not bother to read my post directly above this one?

6. ## Re: trigonometric equations

It's ok now:-) The post came while I was writing. Thank you very much.