Question

How do you solve the identity $\sin \left( x \right) + \sin \left( {2x} \right) = 0$?

Answer 1

Here the given question is based on trigonometric identities and equations. We have to solve the equation for the angle x by using the double angle formula of sine ratio i.e., $\sin \left( {2x} \right) = 2\sin x\cos x$ and further simplify using the standard trigonometric values we get required solution.

Complete step-by-step solution:
Trigonometry the study of the relationships which involve angles, lengths, and heights of triangles. It has many useful identities for learning and deriving the many equations and formulas in science.
Formulas expressing trigonometric functions of an angle $2x$ in terms of functions of an angle $x$,
$\sin 2x = 2\sin x\cos x$

$$\cos 2x = {\cos ^2}x - {\sin ^2}x \\\ \cos 2x = 2{\cos ^2}x - 1 \\\ \cos 2x= 1 - 2{\sin ^2}x \\\$$

$\tan 2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}$
Consider the given trigonometric equation
$\Rightarrow \,\,\,\,\sin \left( x \right) + \sin \left( {2x} \right) = 0$
By sing double angle formula $\sin 2x$ can be written as $2\sin x\cos x$
$\Rightarrow \,\,\,\,\sin \left( x \right) + 2\sin x\cos x = 0$
Take $\sin x$ as common
$\Rightarrow \,\,\,\,\sin \left( x \right)\left( {1 + 2\cos x} \right) = 0$
Equate each factor to the zero, then
$\Rightarrow \,\,\,\,\sin \left( x \right) = 0$ and $1 + 2\cos x = 0$
Consider
$\Rightarrow \,\,\,\,\sin \left( x \right) = 0$
Take inverse sine function on both side, then
$\Rightarrow \,\,\,\,{\sin ^{ - 1}}\left( {\sin x} \right) = {\sin ^{ - 1}}\left( 0 \right)$
As we know $x{x^{ - 1}} = 1$, then
$\Rightarrow \,\,\,\,x = 2n\pi ,\,\,\pi + 2n\pi$, for any integer $n$.
Next, consider
$\Rightarrow \,\,\,1 + 2\cos x = 0$
Take 1 to the RHS
$\Rightarrow \,\,\,2\cos x = - 1$
Divide both side by 2, then
$\Rightarrow \,\,\,\cos x = - \dfrac{1}{2}$
Take inverse cosine function on both side, then
$\Rightarrow \,\,\,{\cos ^{ - 1}}\left( {\cos x} \right) = {\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right)$
$\Rightarrow \,\,\,x = \dfrac{{2\pi }}{3} + 2n\pi ,\,\dfrac{{4\pi }}{3} + 2n\pi$, for any integer $n$.
The final solution is all the values that make $\sin \left( x \right)\left( {1 + 2\cos \left( x \right)} \right) = 0$ true.
$\Rightarrow \,\,\,\,x = 2n\pi ,\pi + 2n\pi ,\dfrac{{2\pi }}{3} + 2n\pi ,\dfrac{{4\pi }}{3} + 2n\pi$
Consolidate $2n\pi$and $\pi + 2n\pi$to $n\pi$.
$x = n\pi ,\dfrac{{2\pi }}{3} + 2n\pi ,\dfrac{{4\pi }}{3} + 2n\pi$ for any integer $n$.

Hence the value of angle x in the trigonometric equation $\sin \left( x \right) + \sin \left( {2x} \right) = 0$ is $x = n\pi ,\dfrac{{2\pi }}{3} + 2n\pi ,\dfrac{{4\pi }}{3} + 2n\pi$ for any integer $n$.

Note: Since we can say at which value of x the addition of cosine and sine will be zero. The ASTC rule defined as all sine tan cosine this explains all trigonometry ratios are positive in the first quadrant. Sine trigonometry ratio is positive in the second quadrant. Tan trigonometry ratio is positive in the third quadrant. The cosine trigonometry ratio is positive in the fourth quadrant.