Number of solutions of sin⁡5x+sin⁡3x+sin⁡x=0 for 0≤x≤π is

sin⁡3x+(sin⁡5x+sin⁡x)=0⇒ sin⁡3x+(2sin⁡3xcos⁡2x)=0⇒ sin⁡3x=0 or cos⁡2x=−12=cos⁡2π3⇒ x=nπ/3 or x=nπ±π/3,n∈Z Then x=0,π/3, and 2π/3, Hence, there are three solutions.

Number of solutions of sin⁡5x+sin⁡3x+sin⁡x=0 for 0≤x≤π is

sin⁡3x+(sin⁡5x+sin⁡x)=0⇒ sin⁡3x+(2sin⁡3xcos⁡2x)=0⇒ sin⁡3x=0 or cos⁡2x=−12=cos⁡2π3⇒ x=nπ/3 or x=nπ±π/3,n∈Z Then x=0,π/3, and 2π/3, Hence, there are three solutions.

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Number of solutions of $\sin 5 x + \sin 3 x + \sin x = 0 for 0 \leq x \leq π$ is

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Detailed Solution

$\sin 3 x + (\sin 5 x + \sin x) = 0$
$\begin{array}{l} \Rightarrow \sin 3 x + (2 \sin 3 xcos 2 x) = 0 \\ \Rightarrow \sin 3 x = 0 or \cos 2 x = - \frac{1}{2} = \cos \frac{2 π}{3} \\ \Rightarrow x = nπ / 3 or x = nπ \pm π / 3, n \in Z \end{array}$
$Then x = 0, π / 3, and 2 π / 3, Hence, there are three solutions.$