Question

Evaluate: $\sin x + \sin 2x = 0$

Answer 1

The given question is a simple trigonometric function $\sin x + \sin 2x = 0$ needs to be solved using the finding up of period values, Formula to find the value of period is given as follows,
The period of the function can be calculated using $\dfrac{{2\pi }}{{\left| b \right|}}$. Then we can get an appropriate solution.

Complete step-by-step answer:
We are given the function, $\sin x + \sin 2x = 0$
Comparing the trigonometric identity, $\sin 2x = \sin x + 2\sin x\cos x$, then
$\Rightarrow \sin (x).1 + 2\sin (x)\cos (x) = 0$
Factor $\sin (x)$ out of $2\sin (x)\cos (x)$
$\Rightarrow \sin (x).1 + \sin (x)(2\cos (x)) = 0$
Factor $\sin (x)$ out of $\sin (x).1 + \sin (x)(2\cos (x))$
$\Rightarrow \sin (x)(1 + 2\cos (x)) = 0$
If any individual factor on the left side of the equation is equal to 0, the entire expression will be equal to 0.

$$\sin (x) = 0 \\\ 1 + 2\cos (x) = 0 \\\$$

Set the first factor equals to 0 and solve.
Set the first factor equal to 0.
$\sin (x) = 0$
Take the inverse sine on both sides of the equation to extract x from inside the sine.
${\sin ^{ - 1}}\sin (x) = {\sin ^{ - 1}}(0)$

$$x = {\sin ^{ - 1}}(0) \\\ x = 0 \\\$$

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from $\pi$to find the solution in the second quadrant.

$$x = \pi - 0 \\\ x = \pi \\\$$

Find the period
The period of the function can be calculated using $\dfrac{{2\pi }}{{\left| b \right|}}$
Replace b with 1 in the formula for period $\dfrac{{2\pi }}{{\left| 1 \right|}}$
The period of the $\sin (x)$function is $2\pi$so values will repeat every $2\pi$radians in both directions.
$x = 2\pi n,\pi + 2\pi n,$for any integer n.
Set the next factor equal to 0 and solve.

$$1 + 2\cos (x) = 0 \\\ 2\cos (x) = - 1 \\\ \cos (x) = \dfrac{{ - 1}}{2} \\\$$

Take the inverse cosine of both sides of the equation to extract x from inside the cosine.

$${\cos ^{ - 1}}\cos (x) = {\cos ^{ - 1}}(\dfrac{{ - 1}}{2}) \\\ x = \dfrac{{2\pi }}{3} \\\$$

The cosine function is negative in the second and third quadrants. To find the second solution, subtract the reference angle from $2\pi$to find the solution in the third quadrant.

$$x = 2\pi - \dfrac{{2\pi }}{3} \\\ x = \dfrac{{6\pi - 2\pi }}{3} \\\ x = \dfrac{{4\pi }}{3} \\\$$

Find the period
The period of the function can be calculated using $\dfrac{{2\pi }}{{\left| b \right|}}$
Replace b with 1 in the formula for period $\dfrac{{2\pi }}{{\left| 1 \right|}}$
The period of the $\cos (x)$function is $2\pi$so values will repeat every $2\pi$radians in both directions.
$x = \dfrac{{2\pi }}{3} + 2\pi n,\dfrac{{4\pi }}{3} + 2\pi n,$for any integer n
The final solution is all the values that make $\sin (x)(1 + 2\cos (x)) = 0$
$x = 2\pi n,\pi + 2\pi n,\dfrac{{2\pi }}{3} + 2\pi n,\dfrac{{4\pi }}{3} + 2\pi n,$for any integer n
Consolidate $2\pi n$ and $\pi + 2\pi n$to $\pi n$,
$x = \pi n,\dfrac{{2\pi }}{3} + 2\pi n,\dfrac{{4\pi }}{3} + 2\pi n,$for any integer n.

Note: We use the trigonometric identity formula to simplify the factor.
Fórmula- $\sin 2x = \sin x + 2\sin x\cos x$
then evaluate the function $\sin x + \sin 2x = 0$.
Factor $\sin (x)$out of $\sin (x) + 2\sin (x)\cos (x)$, Raise $\sin (x)$to the power of 1.
$\sin (x) + 2\sin (x)\cos (x) = 0$
Factor $\sin (x)$out of ${\sin ^1}(x)$
$\sin (x).1 + 2\sin (x)\cos (x) = 0$
Factor $\sin (x)$out of $2\sin (x)\cos (x)$
$\sin (x).1 + \sin (x)(2\cos (x)) = 0$
Factor $\sin (x)$out of $\sin (x).1 + \sin (x)(2\cos (x))$
$\sin (x)(1 + 2\cos (x)) = 0$
The general term of the integer n can be any value which is an infinite, hence it is made consolidated with the value of $2\pi n$ and $\pi + 2\pi n$to $\pi n$, for any integer n.