Question

How do you solve $2{{\sin }^{2}}x=\sin x$?

Answer 1

Hint: We have been given a quadratic equation of $\sin x$. We assume the value of $\sin x$ as the variable $m$. Then we use quadratic solving to solve the problem. We take common terms out to form the multiplied forms. Factorising a polynomial by grouping is to find the pairs which on taking their common divisor out, give the same remaining number. In case of the vanishing method, we use the value of $m$ which gives the polynomial value 0.

Complete answer:
The given equation of $\sin x$ is $2{{\sin }^{2}}x=\sin x$. We assume the term $\sin x$ as the variable $m$.
The revised form of the equation is $2{{m}^{2}}=m$.
We take all the terms in one side and get $2{{m}^{2}}-m=0$
We try to take the common numbers out.
For $2{{m}^{2}}-m$, we take $m$ and get $m\left( 2m-1 \right)$.
The equation becomes $m\left( 2m-1 \right)=0$.
Therefore, $m\left( 2m-1 \right)=0$ has multiplication of two polynomials giving a value of 0. This means at least one of them has to be 0.
So, values of $m$ are $m=0,\dfrac{1}{2}$. This gives $\sin x=0,\dfrac{1}{2}$.
We know that in the principal domain or the periodic value of $-\dfrac{\pi }{2}\le x\le \dfrac{\pi }{2}$ for $\sin x$, if we get $\sin a=\sin b$ where $-\dfrac{\pi }{2}\le a,b\le \dfrac{\pi }{2}$ then $a=b$.
We have $\sin x=0$, the value of $\sin \left( 0 \right)$ as 0. $-\dfrac{\pi }{2}<0<\dfrac{\pi }{2}$.
We have $\sin x=\dfrac{1}{2}$, the value of $\sin \left( \dfrac{\pi }{6} \right)$ as 0. $-\dfrac{\pi }{2}<\dfrac{\pi }{6}<\dfrac{\pi }{2}$.
Therefore, $\sin x=0,\dfrac{1}{2}$ gives $x=0,\dfrac{\pi }{6}$ as primary value.
Therefore, the primary solution for $2{{\sin }^{2}}x=\sin x$ is $x=0,\dfrac{\pi }{6}$ in the domain $-\dfrac{\pi }{2}\le x\le \dfrac{\pi }{2}$.

Note: Although for elementary knowledge the principal domain is enough to solve the problem. But if mentioned to find the general solution then the domain changes to $-\infty \le x\le \infty $. In that case we have to use the formula $x=n\pi +{{\left( -1 \right)}^{n}}a$ for $\sin \left( x \right)=\sin a$ where $-\dfrac{\pi }{2}\le a\le \dfrac{\pi }{2}$. For our given problem $\sin x=0,\dfrac{1}{2}$, the primary solution is $x=0,\dfrac{\pi }{6}$.
The general solution will be $x=\left( n\pi \right)\cup \left( n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{6} \right)$. Here $n\in \mathbb{Z}$.