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Question: How do you solve \(2{{\sin }^{2}}x=\sin x\)?...

How do you solve 2sin⁡2x=sin⁡x2{{\sin }^{2}}x=\sin x?

Explanation

Solution

We have been given a quadratic equation of sin⁡x\sin x. We assume the value of sin⁡x\sin x as the variable mm. 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 mm which gives the polynomial value 0.

Complete answer:
The given equation of sin⁡x\sin x is 2sin⁡2x=sin⁡x2{{\sin }^{2}}x=\sin x. We assume the term sin⁡x\sin x as the variable mm.
The revised form of the equation is 2m2=m2{{m}^{2}}=m.
We take all the terms in one side and get 2m2−m=02{{m}^{2}}-m=0
We try to take the common numbers out.
For 2m2−m2{{m}^{2}}-m, we take mm and get m(2m−1)m\left( 2m-1 \right).
The equation becomes m(2m−1)=0m\left( 2m-1 \right)=0.
Therefore, m(2m−1)=0m\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 mm are m=0,12m=0,\dfrac{1}{2}. This gives sin⁡x=0,12\sin x=0,\dfrac{1}{2}.
We know that in the principal domain or the periodic value of −π2≤x≤π2-\dfrac{\pi }{2}\le x\le \dfrac{\pi }{2} for sin⁡x\sin x, if we get sin⁡a=sin⁡b\sin a=\sin b where −π2≤a,b≤π2-\dfrac{\pi }{2}\le a,b\le \dfrac{\pi }{2} then a=ba=b.
We have sin⁡x=0\sin x=0, the value of sin⁡(0)\sin \left( 0 \right) as 0. −π2<0<π2-\dfrac{\pi }{2}<0<\dfrac{\pi }{2}.
We have sin⁡x=12\sin x=\dfrac{1}{2}, the value of sin⁡(π6)\sin \left( \dfrac{\pi }{6} \right) as 0. −π2<π6<π2-\dfrac{\pi }{2}<\dfrac{\pi }{6}<\dfrac{\pi }{2}.
Therefore, sin⁡x=0,12\sin x=0,\dfrac{1}{2} gives x=0,π6x=0,\dfrac{\pi }{6} as primary value.
Therefore, the primary solution for 2sin⁡2x=sin⁡x2{{\sin }^{2}}x=\sin x is x=0,π6x=0,\dfrac{\pi }{6} in the domain −π2≤x≤π2-\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 −∞≤x≤∞-\infty \le x\le \infty . In that case we have to use the formula x=nπ+(−1)nax=n\pi +{{\left( -1 \right)}^{n}}a for sin⁡(x)=sin⁡a\sin \left( x \right)=\sin a where −π2≤a≤π2-\dfrac{\pi }{2}\le a\le \dfrac{\pi }{2}. For our given problem sin⁡x=0,12\sin x=0,\dfrac{1}{2}, the primary solution is x=0,π6x=0,\dfrac{\pi }{6}.
The general solution will be x=(nπ)∪(nπ+(−1)nπ6)x=\left( n\pi \right)\cup \left( n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{6} \right). Here n∈Zn\in \mathbb{Z}.