If $sin 5 x + sin 3 x + sin x = 0,$ then the value of $x$ other than zero, lying between $0 < x \leq \frac{π}{2}$ is:

\frac{π}{6}

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\frac{π}{12}

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\frac{π}{3}

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\frac{π}{4}

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Solution

The correct option is C $\frac{π}{3}$
$sin 5 x + sin 3 x + sin x = 0$
$(sin 5 x + sin x) + sin 3 x = 0$
$sin C + sin D = 2 sin \frac{C + D}{2} cos \frac{C - D}{2}$
$2 sin 3 x \cdot cos 2 x + sin 3 x = 0$
$sin 3 x (2 cos 2 x + 1) = 0$
$sin 3 x = sin 0$ or $2 cos 2 x + 1 = 0$
$3 x = n π$ $cos 2 x = cos 2 π_{/ 3}$
$x = \frac{n π}{3}$ $2 x = 2 π_{/ 3} + 2 n π$
$0 \leq x \leq π_{/ 2}$ $x = π_{/ 3} \pm n π$ .
$x = n π_{/ 3}$ gives, $x = 0, π_{/ 3}$ .

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If sin5x+sin3x+sinx=0, then the value of x other than zero, lying between 0<x≤π2 is:

The correct option is C π3sin5x+sin3x+sinx=0(sin5x+sinx)+sin3x=0sinC+sinD=2sinC+D2cosC−D22sin3x⋅cos2x+sin3x=0sin3x(2cos2x+1)=0sin3x=sin0 or 2cos2x+1=03x=nπ cos2x=cos2π/3x=nπ3 2x=2π/3+2nπ0≤x≤π/2 x=π/3±nπ.x=nπ/3 gives, x=0,π/3.

If $sin 5 x + sin 3 x + sin x = 0,$ then the value of $x$ other than zero, lying between $0 < x \leq \frac{π}{2}$ is: