If \[\sin 5x + \sin 3x + \sin x = 0\], then the value of \[x\] other than \[0\] lying between \[0 \le x \le \frac{\pi }{2}\]is
A. \[\frac{\pi }{6}\]
B. \[\frac{\pi }{{12}}\]
C. \[\frac{\pi }{3}\]
D. \[\frac{\pi }{4}\]

Hint
In this case, we have been provided in the question that \[\sin 5x + \sin 3x + \sin x = 0\] _{which is lying between}\[0 \le x \le \frac{\pi }{2}\] _{and we have to find the value of} \[x\] other than \[0\] To determine the value, we have to use the formula \[\sin c+\sin d = 2 \sin \left(\frac{c+d}{2}\right) \cos \left(\frac{c-d}{2}\right)\] and by applying the above provided property to the given equation we have to write the equation in the above mentioned form to determine the value of x but we are given that the answer should be other than zero to determine the required solution.

Formula used:
\[\sin c + \sin d = 2\sin \left( {\frac{{c + d}}{2}} \right)\cos \left( {\frac{{c - d}}{2}} \right)\]
Complete step-by-step solution
The given equation is
\[sin5x + \sin 3x + \sin x = 0\]
It can also be solved as,
\[ = > (\sin 5x + \sin x) + \sin 3x = 0\]
By using the trigonometric formula,
\[ = > 2\sin 3x\cos 2x + \sin 3x = 0\]
By solving the equation, it becomes
\[ = > \sin 3x(2\cos 2x + 1) = 0\]
\[ = > \sin 3x = 0;\cos 2x = - 1/2\]
\[ = > 3x = n\pi ,2x=2\pi /3\]
So, the equation is solved as
\[ = 2n\pi \pm 2\pi /3\]
Therefore, the value required is calculated as \[\frac{\pi }{3}\]
Hence, the correct option is C.

Note
The sine, cosine, and tangent ratios are the three main ratios on which all trigonometric functions and formulas are based. The cosine's reciprocal is the secant. In a right triangle, it is the ratio of the hypotenuse to the side next to a particular angle. Because the hypotenuse is 1 on the unit circle,
Because all three sides of a triangle are comparable, multiplying the hypotenuse by any number also multiplies the other sides by that number. The ratio of the perpendicular of a right-angled triangle to the hypotenuse is known as the sine of the angle.