Solve tan(4x) - tan(2x) = 0 for 0 < x < 2π

[xLaser]

Can't figure out this question, a little help would be great.

Solve for X, domain: 0<x<2pi

tan4X - tan2X = 0

What i got is that if we use the double angle formula to expand tan2X into

2tanx / 1-tan^2x

the u move it onto the other side and then somehow tan2x is = to tan4X, I'm very confused, please help.

 

[apchemstudent]

Can't figure out this question, a little help would be great.

Solve for X, domain: 0<x<2pi

tan4X - tan2X = 0

What i got is that if we use the double angle formula to expand tan2X into

2tanx / 1-tan^2x

the u move it onto the other side and then somehow tan2x is = to tan4X, I'm very confused, please help.

for tan 4x you can set it up as tan 2(2x) = 2 tan2x/(1-(tan 2x)^2). Don't simplify tan 2x and just substitute the simplifed form of tan 4x into the equation. The equation is already set to 0 so just solve for x.

Edit: don't forget about the non-permissable values..

 

[dextercioby]

$$\tan 4x=\tan 2x$$ (1)

$$\tan 4x=\frac{2\tan 2x}{1-\tan^{2}2x}$$ (2)

Plug (2) in (1) and solve the eq.Be careful with the 5 points 0,pi/2,pi,3pi/2,2pi.

Daniel.

 

[VietDao29]

Hi,
You can use:
$$\tan 4x=\frac{2\tan 2x}{1-\tan^{2}2x}$$
to solve the problem. Or use this way:
Since you know that: $$\tan{(x + Kpi)} = \tan{x}$$
And $$\tan{4x} = \tan{2x}$$
<=> $$4x - 2x = Kpi$$
<=> $$2x = Kpi$$
<=> $$x = K\frac{pi}{2}$$
K = ...; -3; -2; -1; 0; 1; 2; 3;... (K belongs to Z)
And the rest you can do.
Viet Dao,

 

[xLaser]

ah, i got it, its simply asking for what values of tan2x = 0 in 1 full revolution. Thx.

so technically you could simply do Tan4x - Tan 2X = tan 2X and then tan2x = 0