Question

How do you solve $\sin (2x) + \sin (x) = 0?$

Answer 1

Hint:As we know that the above given question is related to trigonometric expression, sine and cosine are trigonometric ratios. Here we have to prove that the left hand side expression is equal to the right hand expression by using the double angle formula. We know that double angle formula which states that $\sin 2a = 2\sin a\cos a$. We can convert the equation into basic trigonometric equations by applying the trigonometric identities.

Complete step by step solution:
As per the given question we have to solve the equation $\sin (2x) + \sin (x) = 0$. By applying the trigonometric identity i.e. $\sin 2a = 2\sin a\cos a$ in the equation, here $x = a$. So the equation can be written as $2\sin (x)\cos (x) + \sin (x) = 0$. We can see that $\sin (x)$ is common in both the terms, so by taking $\sin (x)$ common we get : $\sin (x)\{ 2\cos (x) + 1\} = 0$.
We get the two possibilities from the above equation, either $\sin (x) = 0$ or $2\cos (x) + 1 = 0$.
Solving for both: If $\sin (x) = 0$ then the value of $x = n\pi $ and if $2\cos (x) + 1 = 0 \Rightarrow x
= - \dfrac{1}{2}$, then the value of $x$ is $2n\pi + \dfrac{{2\pi }}{3},2n\pi - \dfrac{{2\pi }}{3}$.
Hence the required solution of the equation is $x = 2n\pi \pm \dfrac{{2\pi }}{3}$ or $x = n\pi $.

Note: WE should know the values of all trigonometric ratios about from where up to the value lies. We should note that for the trigonometric ratios we have a double angle formula and half angle formula, so by using these formulas, we can solve the trigonometric ratios. The double angle formula for cosine is defined as $\cos (2x) = 2{\cos ^2}x - 1$. Here in the formula $x$ represents the angle.
We can also solve this question by using half angle formulas and later we can use double angle formulas or trigonometric identities to solve further.