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Question: How do you solve \(sin\theta +2sin\theta \cdot \cos \theta =0\) ?...

How do you solve sinθ+2sinθcosθ=0sin\theta +2sin\theta \cdot \cos \theta =0 ?

Explanation

Solution

Problems of solving equations like this can be easily solved by using expressions of θ\theta for equations having trigonometric functions. First, we take the sinθ\sin \theta common and equate the two products to 00 . Now, using the formulas of θ\theta we get the solution of the given problem.

Complete step-by-step answer:
The equation we have is
sinθ+2sinθcosθ=0sin\theta +2sin\theta \cdot \cos \theta =0
We can see that both the terms of the left-hand side of the given equation has sinθ\sin \theta in common.
Hence, we take the term sinθ\sin \theta as common from the left hand-side of the given equation and we get
sinθ(1+2cosθ)=0\Rightarrow sin\theta \left( 1+2\cos \theta \right)=0
We know that if the product of two terms is zero then any one of the terms in the product has to be equal to zero.
Considering the first term sinθ\sin \theta to be zero
sinθ=0\Rightarrow \sin \theta =0
From trigonometric equalities for equation solving, we know that if sinθ=sinφ\sin \theta =\sin \varphi then θ=(1)nφ+nπ\theta ={{\left( -1 \right)}^{n}}\varphi +n\pi where nn is an integer.
As, sin0=0\sin 0=0 we take φ=0\varphi =0
Hence, for sinθ=0\sin \theta =0the expression of θ\theta becomes θ=nπ\theta =n\pi
Again, considering for the term (1+2cosθ)\left( 1+2\cos \theta \right) to be zero we get
1+2cosθ=0\Rightarrow 1+2\cos \theta =0
Subtracting 11 from both the sides of the above equation we get
2cosθ=1\Rightarrow 2\cos \theta =-1
Now, we divide both the sides of the above equation by 22 and get
cosθ=12\Rightarrow \cos \theta =-\dfrac{1}{2}
Again, from trigonometric equalities for equation solving, we know that if cosθ=cosφ\cos \theta =\cos \varphi then θ=2nπ±φ\theta =2n\pi \pm \varphi
As, cos2π3=12\cos \dfrac{2\pi }{3}=-\dfrac{1}{2} and cos4π3=12\cos \dfrac{4\pi }{3}=-\dfrac{1}{2}we get φ=2π3,4π3\varphi =\dfrac{2\pi }{3},\dfrac{4\pi }{3}
Substituting the above expression of φ\varphi in the expression of θ\theta we get
θ=2nπ±2π3\theta =2n\pi \pm \dfrac{2\pi }{3} and θ=2nπ±4π3\theta =2n\pi \pm \dfrac{4\pi }{3}
Therefore, the solutions of the given equation sinθ+2sinθcosθ=0sin\theta +2sin\theta \cdot \cos \theta =0 are θ=nπ, 2nπ±2π3 and 2nπ±4π3\theta =n\pi ,\text{ }2n\pi \pm \dfrac{2\pi }{3}\text{ and }2n\pi \pm \dfrac{4\pi }{3}

Note: While using the expressions of θ\theta we must be careful so that all the alternative solutions are taken into consideration. Also, we must keep in mind that the formulas are used correctly so that flawless solutions are obtained. We should remember that sinθ\sin \theta and cosθ\cos \theta being periodic functions, we should express all the corresponding values by a general solution rather than taking only a single solution unless the domain is mentioned.