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Question: Find the general solution of the trigonometric equation given by, \(\sin x = \tan x\)...

Find the general solution of the trigonometric equation given by, sinx=tanx\sin x = \tan x

Explanation

Solution

To find the general solution we express the function in terms of sin and cos trigonometric functions. We then reduce the equation into a single function and find the solution to it.

Complete step-by-step answer :
Given that,
sinx=tanx\sin x = \tan x ……… (i)
We know that,
tanx=sinxcosx\tan x = \dfrac{{\sin x}}{{\cos x}},
Put this in equation (i),
sinx=sinxcosx\Rightarrow \sin x = \dfrac{{\sin x}}{{\cos x}}
sinxcosx=sinx\Rightarrow \sin x\cos x = \sin x
This can be written as:
cosx=sinxsinx\Rightarrow \cos x = \dfrac{{\sin x}}{{\sin x}}
cosx=1\Rightarrow \cos x = 1
Now, we have to find the value of x, for which cosx=1\cos x = 1
We know that,
cos0=1\cos 0 = 1
Therefore,
cosx=cos0\cos x = \cos 0
We know that,
cosx=cosy\cos x = \cos y, implies x=2nπ±yx = 2n\pi \pm y, where nZn \in Z [Z – set of integers]
Therefore,
cosx=cos0\cos x = \cos 0
Implies,
x=2nπ±0x = 2n\pi \pm 0 or,
x=2nπx = 2n\pi where nZn \in Z [Z is a set of integers]
Hence, the general solution of sinx=tanx\sin x = \tan x is x=2nπx = 2n\pi

Note : In order to solve this type of problems the key is to know the values of angles (x) for some frequent/general values of the function Cos x. We have to remember that for any real numbers x and y, cosx=cosy\cos x = \cos y, implies x=2nπ±yx = 2n\pi \pm y, where nZn \in Z and Z is a set of integers and thus, we get the general solution.