tan x + tan 2x = tan 3x Solve for x?

Answer 1

One way to approach this is to use the double angle identities.

tan(a+b) = [tan(a) + tan(b)]/[1 - tan(a)tan(b)]

We use it by noting that tan(2x) = tan(x+x), and
tan(3x) = tan(2x+x).

Off to the side, let's first calculate tan(2x).

tan(x+x) = [tan(x) + tan(x)]/[1 - tan(x)tan(x)]
tan(2x) = [2tan(x)]/[1 - tan^2(x)]

Now, we calculate.

tan(x) + tan(2x) = tan(3x)
tan(x) + [2tan(x)]/[1 - tan^2(x)] = tan(x + 2x)

Let's put the left hand side under a common denominator.

tan(x)[1 - tan^2(x)] + [2tan(x)]/[1 - tan^2(x)] = tan(x + 2x)

[tan(x) - tan^3(x) + 2tan(x)]/[1 - tan^2(x)] = tan(x + 2x)

Combining like terms,

[3tan(x) - tan^3(x)]/[1 - tan^2(x)] = tan(x + 2x)

Now, let's work out the right hand side.

[3tan(x) - tan^3(x)]/[1 - tan^2(x)] = [tan(x) + tan(2x)]/[1 - tan(x)tan(2x)]

We replace EACH occurance of tan(2x) by what we just solved for, turning it into a mess.

[3tan(x) - tan^3(x)]/[1 - tan^2(x)] = [tan(x) + ([2tan(x)]/[1 - tan^2(x)])]/[1 + tan(x)([2tan(x)]/[1 - tan^2(x)])]

Changing the complex fraction on the side, we'd have to multiply top and bottom by [1 - tan^2(x)], getting

[3tan(x) - tan^3(x)]/[1 - tan^2(x)] = [ tan(x)[1 - tan^2(x)] + 2tan(x) ] / [ (1 - tan^2(x)) + 2tan^2(x) ]

[3tan(x) - tan^3(x)]/[1 - tan^2(x)] = [ tan(x) - tan^3(x) + 2tan(x) ] / [1 + tan^2(x)]

Combining more like terms,

[3tan(x) - tan^3(x)]/[1 - tan^2(x)] = [3tan(x) - tan^3(x)]/[1+tan^2(x)]

Now that we have two simple fractions we cross multiply.

[3tan(x) - tan^3(x)][1+tan^2(x)] = [3tan(x) - tan^3(x)][1 - tan^2(x)]

Expanding these binomials,

3tan(x) + 2tan^3(x) - tan^3(x) - tan^5(x) = 3tan(x) - 3tan^3(x) - tan^3(x) + tan^5(x)

Bringing everything over to the left hand side, we get

3tan(x) + 2tan^3(x) - tan^3(x) - tan^5(x) - 3tan(x) + 3tan^3(x) + tan^3(x) - tan^5(x) = 0

Combine like terms

5tan^3(x) -2tan^5(x) = 0

Factor out a tan^3(x), to get

tan^3(x) [5 - 2tan^2(x)] = 0

Now, equate each to 0

tan^3(x) = 0
5 - 2tan^2(x) = 0

And solve individually.

tan^3(x) = 0, implies
tan(x) = 0, so x = 0, pi

5 - 2tan^2(x) = 0, implies
tan^2(x) = 5/2
Therefore,
tan(x) = sqrt(5/2)

To solve this you need to take the arctan of sqrt(5/2)

Answer 2

x=1

Answer 3

3x=2x+x utilizing tan b/s tan3x=tan(2x+x) tan3x=tan2x+tanx1-tan2xtanx tan3x(a million-tan2xtanx)=tan2x+tanx tan3x-tan3xtan2xtanx=tan2x+tanx tan3x-tan2x-tanx=tan3xtan2xtanx L.H.S=R.H.S