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Question: Find the general solution of \(sinx=tanx\)....

Find the general solution of sinx=tanxsinx=tanx.

Explanation

Solution

Hint: Use the fact that if sin⁡x=sin⁡y\sin x=\sin y, then x=nπ+(−1)ny,n∈Zx=n\pi +{{\left( -1 \right)}^{n}}y,n\in \mathbb{Z} and if cos⁡x=cos⁡y\cos x=\cos y, then x=2nπ±y,n∈Zx=2n\pi \pm y,n\in \mathbb{Z}. Use sin⁡xcos⁡x=tan⁡x\dfrac{\sin x}{\cos x}=\tan x and write the whole equation in terms of sines and cosines. Factorise the expression and use zero product property. Also, keep in mind that cos⁡x≠0\cos x\ne 0. Hence write the general solution of the equation.

Complete step-by-step answer:
We have sinx = tanx.
We know that tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x}
Hence, we have
sin⁡x=sin⁡xcos⁡x\sin x=\dfrac{\sin x}{\cos x}
Multiplying both sides by cosx, we get
sin⁡xcos⁡x=sin⁡x\sin x\cos x=\sin x
Subtracting sin(x) from both sides, we get
sin⁡xcos⁡x−sin⁡x=0\sin x\cos x-\sin x=0
Taking sin(x) common from the terms in LHS, we get
sin⁡x(cos⁡x−1)=0\sin x\left( \cos x-1 \right)=0
Using zero product property, we get
sin⁡x=0\sin x=0 or cos⁡x−1=0\cos x-1=0
Solving sinx = 0:
We have sin(0) = 1
Hence sin(x) = sin(0)
We know that the general solution of the equation sin⁡x=sin⁡y\sin x=\sin y is given by x=nπ+(−1)ny,n∈Zx=n\pi +{{\left( -1\right)}^{n}}y,n\in \mathbb{Z}
Hence we have
x=nπ+(−1)n(0)=nπ,n∈Zx=n\pi +{{\left( -1 \right)}^{n}}\left( 0 \right)=n\pi ,n\in \mathbb{Z}
Solving cosx -1 =0:
We have cosx = 1
We know that cos⁡(0)=1\cos \left( 0 \right)=1
Hence, we have
cos⁡x=cos⁡(0)\cos x=\cos \left( 0 \right)
We know that the general solution of the equation cosx = cosy is given by x=2nπ±yx=2n\pi \pm y
Hence we have
x=2nπ±0=2nπ,n∈Zx=2n\pi \pm 0=2n\pi ,n\in \mathbb{Z}
Also, observe that \forall x\in \left\\{ n\pi ,n\in \mathbb{Z} \right\\},\cos x\ne 0 and \forall x\in\left\\{ 2n\pi ,n\in \mathbb{Z} \right\\},\cos x\ne 0
Hence we have
Hence we have x\in \left\\{ n\pi ,n\in \mathbb{Z} \right\\}\bigcup \left\\{ 2n\pi ,n\in \mathbb{Z} \right\\}

Observe that \left\\{ 2n\pi ,n\in \mathbb{Z} \right\\}\subset \left\\{ n\pi ,n\in \mathbb{Z} \right\\}
\left\\{ n\pi ,n\in \mathbb{Z} \right\\}\bigcup \left\\{ 2n\pi ,n\in \mathbb{Z} \right\\}=\left\\{ n\pi ,n\in \mathbb{Z} \right\\}
Hence x\in \left\\{ n\pi ,n\in \mathbb{Z} \right\\}, which is the general solution of the given equation sinx = tanx.

Note: Whenever solving equations involving tanx,cotx, secx and cosecx , do not forget to remove the
points in the domain at which they are undefined. This mistake is made by many students while
solving trigonometric equations.