Question
Question: How do you solve \(sin\theta +2sin\theta \cdot \cos \theta =0\) ?...
How do you solve sinθ+2sinθ⋅cosθ=0 ?
Solution
Problems of solving equations like this can be easily solved by using expressions of θ for equations having trigonometric functions. First, we take the sinθ common and equate the two products to 0 . Now, using the formulas of θ we get the solution of the given problem.
Complete step-by-step answer:
The equation we have is
sinθ+2sinθ⋅cosθ=0
We can see that both the terms of the left-hand side of the given equation has sinθ in common.
Hence, we take the term sinθ as common from the left hand-side of the given equation and we get
⇒sinθ(1+2cosθ)=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 to be zero
⇒sinθ=0
From trigonometric equalities for equation solving, we know that if sinθ=sinφ then θ=(−1)nφ+nπ where n is an integer.
As, sin0=0 we take φ=0
Hence, for sinθ=0the expression of θ becomes θ=nπ
Again, considering for the term (1+2cosθ) to be zero we get
⇒1+2cosθ=0
Subtracting 1 from both the sides of the above equation we get
⇒2cosθ=−1
Now, we divide both the sides of the above equation by 2 and get
⇒cosθ=−21
Again, from trigonometric equalities for equation solving, we know that if cosθ=cosφ then θ=2nπ±φ
As, cos32π=−21 and cos34π=−21we get φ=32π,34π
Substituting the above expression of φ in the expression of θ we get
θ=2nπ±32π and θ=2nπ±34π
Therefore, the solutions of the given equation sinθ+2sinθ⋅cosθ=0 are θ=nπ, 2nπ±32π and 2nπ±34π
Note: While using the expressions of θ 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 and cosθ 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.