
Find the general solution of $sinx=tanx$.
Answer
600.9k+ views
Hint: Use the fact that if $\sin x=\sin y$, then $x=n\pi +{{\left( -1 \right)}^{n}}y,n\in \mathbb{Z}$ and if $\cos x=\cos y$, then $x=2n\pi \pm y,n\in \mathbb{Z}$. Use $\dfrac{\sin x}{\cos x}=\tan x$ and write the whole equation in terms of sines and cosines. Factorise the expression and use zero product property. Also, keep in mind that $\cos x\ne 0$. Hence write the general solution of the equation.
Complete step-by-step answer:
We have sinx = tanx.
We know that $\tan x=\dfrac{\sin x}{\cos x}$
Hence, we have
\[\sin x=\dfrac{\sin x}{\cos x}\]
Multiplying both sides by cosx, we get
$\sin x\cos x=\sin x$
Subtracting sin(x) from both sides, we get
$\sin x\cos x-\sin x=0$
Taking sin(x) common from the terms in LHS, we get
$\sin x\left( \cos x-1 \right)=0$
Using zero product property, we get
$\sin x=0$ or $\cos x-1=0$
Solving sinx = 0:
We have sin(0) = 1
Hence sin(x) = sin(0)
We know that the general solution of the equation $\sin x=\sin y$ is given by $x=n\pi +{{\left( -1\right)}^{n}}y,n\in \mathbb{Z}$
Hence we have
$x=n\pi +{{\left( -1 \right)}^{n}}\left( 0 \right)=n\pi ,n\in \mathbb{Z}$
Solving cosx -1 =0:
We have cosx = 1
We know that $\cos \left( 0 \right)=1$
Hence, we have
$\cos x=\cos \left( 0 \right)$
We know that the general solution of the equation cosx = cosy is given by $x=2n\pi \pm y$
Hence we have
$x=2n\pi \pm 0=2n\pi ,n\in \mathbb{Z}$
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.
Complete step-by-step answer:
We have sinx = tanx.
We know that $\tan x=\dfrac{\sin x}{\cos x}$
Hence, we have
\[\sin x=\dfrac{\sin x}{\cos x}\]
Multiplying both sides by cosx, we get
$\sin x\cos x=\sin x$
Subtracting sin(x) from both sides, we get
$\sin x\cos x-\sin x=0$
Taking sin(x) common from the terms in LHS, we get
$\sin x\left( \cos x-1 \right)=0$
Using zero product property, we get
$\sin x=0$ or $\cos x-1=0$
Solving sinx = 0:
We have sin(0) = 1
Hence sin(x) = sin(0)
We know that the general solution of the equation $\sin x=\sin y$ is given by $x=n\pi +{{\left( -1\right)}^{n}}y,n\in \mathbb{Z}$
Hence we have
$x=n\pi +{{\left( -1 \right)}^{n}}\left( 0 \right)=n\pi ,n\in \mathbb{Z}$
Solving cosx -1 =0:
We have cosx = 1
We know that $\cos \left( 0 \right)=1$
Hence, we have
$\cos x=\cos \left( 0 \right)$
We know that the general solution of the equation cosx = cosy is given by $x=2n\pi \pm y$
Hence we have
$x=2n\pi \pm 0=2n\pi ,n\in \mathbb{Z}$
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.
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