Question

Tan theta + tan 2 theta + root 3 tan 2 theta tan theta equals to root 3 find the general solution

Answer 1

Given: tan Ф + tan 2Ф + √3 x tan Ф x tan 2Ф = √3

To find:  The general solution?

Solution:

• Now we have given the trigonometric equation as:

            tan Ф + tan 2Ф + √3 x tan Ф x tan 2Ф = √3

• Now simplifying it, we get:

            tan Ф + tan 2Ф = √3 - √3 x tan Ф x tan 2Ф

            tan Ф + tan 2Ф = √3 ( 1 -  tan Ф x tan 2Ф )

            tan Ф + tan 2Ф / 1 - tan Ф x tan 2Ф = √3

• Now we know that tan( a+ b) = tan a + tan b / 1 - tan a tan b, so:

            tan ( Ф + 2Ф ) = √3

            tan 3Ф = √3

            tan 3Ф = tan π/3

            3Ф = π/3

            Ф = π/9 or nπ/3 + π/9

           Ф = π/9 ( 3n + 1 )    n ∈ Z

Answer:

        So the general solution of  tan Ф + tan 2Ф + √3 x tan Ф x tan 2Ф = √3 is π/9 ( 3n + 1 ).

Answer 2

Answer:

$\tan(θ) + \tan(2θ) + \sqrt{3} tan(2θ)tan(θ) = \sqrt{3}$

$\tanθ + \tan(2θ) = \sqrt{3} (1 - tanθtan(2θ))$

$\frac{(tan(θ) +tan(2θ)) }{(1 - tan(2θ)tan(θ))} = \sqrt{3}$

$tan(3θ) = \sqrt{3} = tan \frac{\pi}{3}$

$(3θ) = n\pi + \frac{\pi}{3}$

$(3θ) = n\pi + \frac{\pi}{3} (nϵz)$

$θ = \frac{n\pi}{3} + \frac{\pi}{9} (nϵZ)$