
How do you solve the given trigonometric equation ${{\sin }^{2}}x-2\sin x=0$?
Answer
563.4k+ views
Hint: We start solving the problem by factorizing the given trigonometric equation and then equating each of the obtained factors to zero. We then make use of the fact that the range of values of sine function as $\left[ -1,1 \right]$ to neglect the impossible results. We then make use of the fact that that if $\sin x=0$, then the general solution of x is defined as $x=n\pi $ to get the required answer for the given problem.
Complete step by step answer:
According to the problem, we are asked to solve the given trigonometric equation ${{\sin }^{2}}x-2\sin x=0$.
We have given ${{\sin }^{2}}x-2\sin x=0$.
Let us factorize the given trigonometric equation.
$\Rightarrow \left( \sin x\times \sin x \right)+\left( \sin x\times -2 \right)=0$ ---(1).
From equation (1), we can see that both the terms have a common factor $\sin x$. So, let us take that factor out to get the result of factorization.
$\Rightarrow \sin x\left( \sin x-2 \right)=0$.
So, we have \[\sin x=0\], $\sin x-2=0$.
$\Rightarrow \sin x=0$, $\sin x=2$.
We know that the range of values of sine function is $\left[ -1,1 \right]$, which means that sine function cannot be equal to 2. So, we neglect the possibility of $\sin x=2$.
Now, we have $\sin x=0$ ---(2).
We know that if $\sin x=0$, then the general solution of x is defined as $x=n\pi $, $n\in Z$.
So, the required general solution is $x=n\pi $, $n\in Z$.
$\therefore $ We have found the general solution for the given trigonometric equation ${{\sin }^{2}}x-2\sin x=0$ as $x=n\pi $, $n\in Z$.
Note:
Whenever we get this type of problem, we first try to factorize the given trigonometric equation to get the required solution. Here we are assumed that we need to find the general solution of the given trigonometric solution otherwise the result will be different. We should not forget to check the range of the trigonometric functions while solving this type of problem as this may give incorrect results. Similarly, we can expect problems to find the particular solution of the trigonometric equation ${{\tan }^{2}}x+4\tan x+3=0$.
Complete step by step answer:
According to the problem, we are asked to solve the given trigonometric equation ${{\sin }^{2}}x-2\sin x=0$.
We have given ${{\sin }^{2}}x-2\sin x=0$.
Let us factorize the given trigonometric equation.
$\Rightarrow \left( \sin x\times \sin x \right)+\left( \sin x\times -2 \right)=0$ ---(1).
From equation (1), we can see that both the terms have a common factor $\sin x$. So, let us take that factor out to get the result of factorization.
$\Rightarrow \sin x\left( \sin x-2 \right)=0$.
So, we have \[\sin x=0\], $\sin x-2=0$.
$\Rightarrow \sin x=0$, $\sin x=2$.
We know that the range of values of sine function is $\left[ -1,1 \right]$, which means that sine function cannot be equal to 2. So, we neglect the possibility of $\sin x=2$.
Now, we have $\sin x=0$ ---(2).
We know that if $\sin x=0$, then the general solution of x is defined as $x=n\pi $, $n\in Z$.
So, the required general solution is $x=n\pi $, $n\in Z$.
$\therefore $ We have found the general solution for the given trigonometric equation ${{\sin }^{2}}x-2\sin x=0$ as $x=n\pi $, $n\in Z$.
Note:
Whenever we get this type of problem, we first try to factorize the given trigonometric equation to get the required solution. Here we are assumed that we need to find the general solution of the given trigonometric solution otherwise the result will be different. We should not forget to check the range of the trigonometric functions while solving this type of problem as this may give incorrect results. Similarly, we can expect problems to find the particular solution of the trigonometric equation ${{\tan }^{2}}x+4\tan x+3=0$.
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