Solve for x where 0 <= x < 2pi:
tan(2x) * tan(4x) = 1
tan(4x) = tan(2*(2x)) = 2*tan(2x)/[1 - tan^2(2x)]
So
tan(2x) * tan(4x) = 1
becomes
2*tan^2(2x)/[1 - tan^2(2x)] = 1
2*tan^2(2x) = 1 - tan^2(2x)
3*tan^2(2x) = 1
tan^2(2x) = 1/3
tan(2x) = 1/sqrt{3} <-- Only keep the + solution since 0 <= x < 2(pi)
2x = (pi)/6 (rad)
x = (pi)/12 (rad)
-Dan
Yes, I didn't give you the full solution set, did I?
Okay, let's do this the hard way:
tan(2x)*tan(4x) = 1
tan(2x) * [2*tan(2x)/(1 - tan^2(2x))] = 1
2*tan^2(2x)/[1 - tan^2(2x)] = 1
2*tan^2(2x) = 1 - tan^2(2x)
3*tan^2(2x) = 1
So far so good. Now let's expand tan(2x):
3*[2*tan(x)/(1 - tan^2(x))]^2 = 1
12*tan^2(x)/(1 - tan^2(x))^2 = 1
12*tan^2(x) = 1 - 2*tan^2(x) + tan^4(x)
tan^4(x) - 14*tan^2(x) + 1 = 0
Let y = tan^2(x). Then:
y^2 - 14y + 1 = 0
y = 7 (+/-) 4*sqrt{3}
So
tan^2(x) = 7 (+/-) 4*sqrt{3}
or
tan(x) = (+/-)sqrt{7 + 4*sqrt{3}} or tan(x) = (+/-)sqrt{7 - 4*sqrt{3}}
Where to go from here? Well:
7 + 4*sqrt{3} = (2 + sqrt{3})^2, so
tan(x) = (+/-)(2 + sqrt{3}) (With a small side of thanks to a recent post of Soroban's )
and 7 - 4*sqrt{3} = (2 - sqrt{3})^2, so
tan(x) = (+/-)(2 - sqrt{3})
This generates the solution set:
x = (+/-)5(pi)/12, (+/-)(pi)/12
But we want 0 <= x < 2(pi) so these become:
x = (pi)/12, 5(pi)/12, 19(pi)/12, and 23(pi)/12
How do we get the others? Well we may add (pi) to any solution we get from arctangent, so we get the additional solutions:
x = 13(pi)/12, 17(pi)/12, 31(pi)/12, and 35(pi)/12
These last two are greater than 2(pi), so we subtract 2(pi) to get
x = 13(pi)/12, 17(pi)/12, 7(pi)/12, and 11(pi)/12
Between the two sets of values for x, we have your full solution.
I guess I shouldn't have taken the short-cut in my first solution.
-Dan
Hello, Mr_Green!
Solve for x where 0 < x < 2π
. . tan(2x)·tan(4x) .= .1
Using a double-angle identity, we have;
. . . . . . . . . . 2·tan(2x)
. . tan(2x) . -------------- · = . 1
. . . . . . . . .1 - tanČ(2x)
Then: .2·tanČ(2x) .= .1 - tanČ(2x) . . → . . 3·tanČ(2x) .= .1
. . . . . . . - . . . - . . . . - . . . . . . . . . . . . . . . . . . . . . . _
And we have: .tanČ(2x) .= .1/3 . . → . . tan(2x) .= .±1/√3
Hence: .2x .= .π/6, 5π/6, 7π/6, 11π/6, 13π/6, 17π/6, 19π/6, 23π/6
Therefore: .x .= .π/12, 5π/12, 7π/12, 11π/12, 13π/12, 17π/12, 19π/12, 23π/12