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Question: How do you solve: \(2{{\sin }^{2}}\theta -\sin \theta =0\)?...

How do you solve: 2sin⁡2θ−sin⁡θ=02{{\sin }^{2}}\theta -\sin \theta =0?

Explanation

Solution

We have been given a quadratic equation of sin⁡θ\sin \theta . We assume the value of sin⁡θ\sin \theta 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, gives the same remaining number. In the case of the vanishing method, we use the value of mm which gives the polynomial value 0.

Complete step by step answer:
The given equation of sin⁡x\sin x is 2sin⁡2θ−sin⁡θ=02{{\sin }^{2}}\theta -\sin \theta =0. We assume the term sin⁡θ\sin \theta as the variable mm
The revised form of the equation is 2m2−m=02{{m}^{2}}-m=0.
We try to take the common numbers out. 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⁡θ=0,12\sin \theta =0,\dfrac{1}{2}.
We know that in the principal domain or the periodic value of −π2≤θ≤π2-\dfrac{\pi }{2}\le \theta \le \dfrac{\pi }{2} for sin⁡θ\sin \theta , 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⁡θ=0\sin \theta =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⁡θ=12\sin \theta =\dfrac{1}{2}, the value of sin⁡(π6)\sin \left( \dfrac{\pi }{6} \right) as 12\dfrac{1}{2}. −π2<π6<π2-\dfrac{\pi }{2}<\dfrac{\pi }{6}<\dfrac{\pi }{2}.
Therefore, sin⁡θ=0,12\sin \theta =0,\dfrac{1}{2} gives θ=0,π6\theta =0,\dfrac{\pi }{6} as primary value.
Therefore, the primary solution for 2sin⁡2θ−sin⁡θ=02{{\sin }^{2}}\theta -\sin \theta =0 is θ=0,π6\theta =0,\dfrac{\pi }{6} in the domain −π2≤θ≤π2-\dfrac{\pi }{2}\le \theta \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⁡θ=0,12\sin \theta =0,\dfrac{1}{2}, the primary solution is θ=0,π6\theta =0,\dfrac{\pi }{6}.
The general solution will be θ=(nπ)∪(nπ+(−1)nπ6)\theta =\left( n\pi \right)\cup \left( n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{6} \right). Here n∈Zn\in \mathbb{Z}.