Question

How do you solve \[\sin (2t) + \sin (t) = 0\]?

Answer 1

Hint: First, we know that double angle identities are special cases of the sum identities. That is, when the two angles are equal, the sum identities are reduced to double angle identities.
The double sine formula is,
$\sin 2A = 2\sin A\cos A$
We will substitute this double angle formula in the given equation. Then, we will take sin common and keep each factor equal to 0.
After we solve for $\cos t$ and find the value of $t$
And finally, we will find the value of $t$.
$\sin t$ will repeat after every pie, so we find $\pi $ and $2\pi $.

Complete step-by-step solution:
The given equation is, \[\sin (2t) + \sin (t) = 0\]
We know that$\sin 2A = 2\sin A\cos A$ .
Convert $A$ into $t$, hence we get,
$ \Rightarrow \sin (2t) = 2\sin t\cos t$
Putting this in the given equation,
$ \Rightarrow 2\sin t\cos t + \sin t = 0$
Now we will take $\sin t$ common in LHS (Left Hand Side), hence we get
$ \Rightarrow \sin t(2\cos t + 1) = 0$
If any individual factor on the left side of the equation is equal to $0$, the entire expression will be equal to $0$.
We separate the sine and cosine, we get
$ \Rightarrow \sin t = 0$ or
$ \Rightarrow 2\cos t + 1 = 0$
Now we will find the value of $\cos t$, interchange LHS to RHS, we get
$ \Rightarrow 2\cos t = - 1$
Dividing by $2$ on both sides, we get
$ \Rightarrow \cos t = - \dfrac{1}{2}$
Now find the value of $t$
$ \Rightarrow t = co{s^{ - 1}}\left( { - \dfrac{1}{2}} \right)$
The exact value of $co{s^{ - 1}}\left( { - \dfrac{1}{2}} \right)$ is $\dfrac{{2\pi }}{3}$ and it can also be $\dfrac{{4\pi }}{3}$ .
Since we are not given any range within which our \[t\] lies, we will have to find a general solution.
If $\cos \theta = \cos \beta $ , then $\theta = 2n\pi + \beta $ . Using this, we get,
$t = 2n\pi + \dfrac{{2\pi }}{3}$ and $t = 2n\pi + \dfrac{{4\pi }}{3}$
Now, set the first factor equal to $0$.
$ \Rightarrow \sin t = 0$
Take the inverse function of both sides of the equation to extract $t$ from inside the sine
$ \Rightarrow t = si{n^{ - 1}}(0)$
The exact value of $si{n^{ - 1}}(0)$ is $0$ but it can also be $\pi $ . Since we are not given any range within which our t lies, we will have to find a general solution.
If $\sin \theta = \sin \alpha $ , then, $\theta = n\pi + \alpha $ . Using this,
$\sin t = 0 = \sin 0^\circ $ , $t = n\pi + 0 = n\pi $ .
Therefore, we have got the required values as $n\pi ,2n\pi + \dfrac{{2\pi }}{3},2n\pi + \dfrac{{4\pi }}{3}$ .
The final solution is all the values that make,
$ \Rightarrow t = 2\pi n,\pi + 2\pi n,\dfrac{{2\pi }}{3} + 2\pi n,\dfrac{{4\pi }}{3} + 2\pi n$, for any integer$n$.
Consolidate $2\pi n$ and $\pi + 2\pi n$ to $\pi n$

$ t = \pi n,\dfrac{{2\pi }}{3} + 2\pi n,\dfrac{{4\pi }}{3} + 2\pi n$ , for any integer $n$.

Note: Application of $\sin 2A = 2\sin A\cos A$: When an object is projected with speed $u$ at an angle $\alpha $to the horizontal over level, the horizontal distance (Range) it travels before striking the ground is given by the formula
$R = \dfrac{{{u^2}\sin 2\alpha }}{g}$
Clearly maximum of $R$ is $\dfrac{{{u^2}}}{g}$, when$\alpha = \dfrac{\pi }{4}$