Solveeit Logo
Login

Question

Question: How do you solve \( {\tan ^2}x = \tan x \) ?...

How do you solve tan⁡2x=tan⁡x{\tan ^2}x = \tan x ?

Explanation

Solution

Hint : Every trigonometric function and formulae are designed on the basis of three primary ratios. Sine, Cosine and tangents are these ratios in trigonometry based on Perpendicular, Hypotenuse and Base of a right triangle . In order to calculate the angles sin , cos and tan functions . According to the formula of tan = sin⁡θcos⁡θ\dfrac{{\sin \theta }}{{\cos \theta }} , we will find the value of x . Also we will find x by transposing as required by the question.

Complete step-by-step answer :
We are given the question as tan⁡2x=tan⁡x{\tan ^2}x = \tan x , we will subtract the function tan x from both the sides L. H. S. and R. H. S.
tan⁡2x−tan⁡x=tan⁡x−tan⁡x{\tan ^2}x - \tan x = \tan x - \tan x
In R. H. S. there it is left with zero , so that we can solve –
tan⁡2x−tan⁡x=0{\tan ^2}x - \tan x = 0
tan⁡x(tan⁡x−1)=0\tan x(\tan x - 1) = 0
Here we are taking the common from the L. H. S. to make it simpler and value of x can be determined ,
Like the quadratic equations we can solve the x as ,
According to the formula of tan = sin⁡θcos⁡θ\dfrac{{\sin \theta }}{{\cos \theta }} , we will find the value of x as tan x can be equated with zero and the tan x - 1 can be equated with zero separately .
tan⁡x=0and tan⁡x−1=0 tan⁡x=1   \tan x = 0 and \\\ \tan x - 1 = 0 \\\ \tan x = 1 \;
According to the formula of tan = sin⁡θcos⁡θ\dfrac{{\sin \theta }}{{\cos \theta }} ,
sin⁡xcos⁡x=0\dfrac{{\sin x}}{{\cos x}} = 0 or sin⁡θcos⁡θ=1\dfrac{{\sin \theta }}{{\cos \theta }} = 1
Now , we will multiply by cos x on both the sides L. H. S. and R. H. S. , we get -
sin⁡x=0\sin x = 0 or sin⁡x=cos⁡x\sin x = \cos x
Now we have to calculate for angle x for which we will give a general solution ,
x=kπx = k\pi or x=π4+kπx = \dfrac{\pi }{4} + k\pi where k∈Zk \in \mathbb{Z}
This is the final answer .
So, the correct answer is “ x=kπx = k\pi or x=π4+kπx = \dfrac{\pi }{4} + k\pi”.

Note : Even Function – A function f(x)f(x) is said to be an even function ,if f(−x)=f(x)f( - x) = f(x) for all x in its domain.
Odd Function – A function f(x)f(x) is said to be an even function ,if f(−x)=−f(x)f( - x) = - f(x) for all x in its domain.
Periodic Function= A function f(x)f(x) is said to be a periodic function if there exists a real number T > 0 such that f(x+T)=f(x)f(x + T) = f(x) for all x.
If T is the smallest positive real number such that f(x+T)=f(x)f(x + T) = f(x) for all x, then T is called the fundamental period of f(x)f(x) .
Since sin⁡ (2nπ+θ)=sin⁡θ\sin \,(2n\pi + \theta ) = \sin \theta for all values of θ\theta and n ∈\in N.
It should be very clear that sin(A+B)sin\left( {A + B} \right)is not equal to sin⁡A+sin⁡B\sin A + \sin B .