Question

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

Answer 1

Hint: Problems of solving equations like this can be easily solved by using expressions of $\theta $ for equations having trigonometric functions. First, we take the $\sin \theta $ common and equate the two products to $0$ . Now, using the formulas of $\theta $ we get the solution of the given problem.

Complete step-by-step answer:
The equation we have is
$sin\theta +2sin\theta \cdot \cos \theta =0$
We can see that both the terms of the left-hand side of the given equation has $\sin \theta $ in common.
Hence, we take the term $\sin \theta $ as common from the left hand-side of the given equation and we get
$\Rightarrow sin\theta \left( 1+2\cos \theta \right)=0$
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 $\sin \theta $ to be zero
$\Rightarrow \sin \theta =0$
From trigonometric equalities for equation solving, we know that if $\sin \theta =\sin \varphi $ then \[\theta ={{\left( -1 \right)}^{n}}\varphi +n\pi \] where $n$ is an integer.
As, $\sin 0=0$ we take $\varphi =0$
Hence, for $\sin \theta =0$the expression of $\theta $ becomes \[\theta =n\pi \]
Again, considering for the term $\left( 1+2\cos \theta \right)$ to be zero we get
$\Rightarrow 1+2\cos \theta =0$
Subtracting $1$ from both the sides of the above equation we get
$\Rightarrow 2\cos \theta =-1$
Now, we divide both the sides of the above equation by $2$ and get
$\Rightarrow \cos \theta =-\dfrac{1}{2}$
Again, from trigonometric equalities for equation solving, we know that if $\cos \theta =\cos \varphi $ then \[\theta =2n\pi \pm \varphi \]
As, $\cos \dfrac{2\pi }{3}=-\dfrac{1}{2}$ and $\cos \dfrac{4\pi }{3}=-\dfrac{1}{2}$we get $\varphi =\dfrac{2\pi }{3},\dfrac{4\pi }{3}$
Substituting the above expression of $\varphi $ in the expression of $\theta $ we get
\[\theta =2n\pi \pm \dfrac{2\pi }{3}\] and \[\theta =2n\pi \pm \dfrac{4\pi }{3}\]
Therefore, the solutions of the given equation $sin\theta +2sin\theta \cdot \cos \theta =0$ are \[\theta =n\pi ,\text{ }2n\pi \pm \dfrac{2\pi }{3}\text{ and }2n\pi \pm \dfrac{4\pi }{3}\]

Note: While using the expressions of $\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 $\sin \theta $ and $\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.