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Question: How does one find the general solution for \(\tan x - 3\cot x = 0\) ?...

How does one find the general solution for tanx3cotx=0\tan x - 3\cot x = 0 ?

Explanation

Solution

For the given equation firstly convert the cot\cot in terms of tan\tan . Then simplify by taking the square root of both sides. Lastly used trigonometric identities and general solution formula for,
tanθ=tanα, θ=α+kπ\tan \theta = \tan \alpha , \\\ \Rightarrow \theta = \alpha + k\pi
Where,kk is any integer value.

Complete step by step answer:
We have the following equation ,
tanx3cotx=0\tan x - 3\cot x = 0 ,
Now by using the trigonometric property i.e., cotx=1tanx\cot x = \dfrac{1}{{\tan x}} ,
Convert the cot\cot in terms of tan\tan .
We get ,
tanx3tanx=0 tan2x3=0\Rightarrow \tan x - \dfrac{3}{{\tan x}} = 0 \\\ \Rightarrow {\tan ^2}x - 3 = 0
Where tanx0\tan x \ne 0 ,
After simplifying ,
tan2x=3 tanx=±3\Rightarrow {\tan ^2}x = 3 \\\ \Rightarrow \tan x = \pm \sqrt 3
Now we have two values , one with positive sign and another with negative sign.
For tanx=3\tan x = \sqrt 3 ,
x=π3\Rightarrow x = \dfrac{\pi }{3}
For this the general solution is
x=π3+kπ\Rightarrow x = \dfrac{\pi }{3} + k\pi
Where, kkis any integer value .
And for tanx=3\tan x = - \sqrt 3 ,
x=2π3\Rightarrow x = \dfrac{{2\pi }}{3}
For this the general solution is
x=2π3+kπ\therefore x = \dfrac{{2\pi }}{3} + k\pi
Where ,kk is any integer value.

Additional Information:
Trigonometric equations can have two types of solutions one is principal solution and other is general solution. Principal solution is the one where the equation involves a variable 0x2π0 \leqslant x \leqslant 2\pi . General solution is the one which involves an integer, say kk and provides all the solutions of the trigonometric equation.

Note: You must know that sin, cosine, and tangent are considered as major trigonometric functions, hence we can derive the solutions for the equations comprising these trigonometric functions or ratios. We can also derive the solutions for the other three trigonometric functions such as secant, cosecant, and cotangent with the help of the solutions which are already derived.