Question

sin 2x - sin 3x = 0 ,find general solutions​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:sin2x - sin3x = 0$

can be rewritten as

$\rm :\longmapsto\:sin3x - sin2x = 0$

We know,

$\boxed{ \sf \: sinx - siny = 2cos\bigg(\dfrac{x + y}{2} \bigg)sin\bigg(\dfrac{x - y}{2} \bigg)}$

So, using this identity we get

$\rm :\longmapsto\:2cos\bigg(\dfrac{3x + 2x}{2} \bigg)sin\bigg(\dfrac{3x - 2x}{2} \bigg) = 0$

$\rm :\longmapsto\:cos\bigg(\dfrac{5x}{2} \bigg)sin\bigg(\dfrac{x}{2} \bigg) = 0$

$\rm :\implies\:cos\bigg(\dfrac{5x}{2} \bigg) = 0 \: \: \: or \: \: \: \: sin\bigg(\dfrac{x}{2} \bigg) = 0$

$\rm :\implies\:\dfrac{5x}{2} = (2n + 1)\dfrac{\pi}{2} \: \: \: or \:\dfrac{x}{2} = n\pi \: \: where \: n \in \: Z$

$\{ \because \: \sf \: cosx = 0 \implies \: x = (2n + 1)\dfrac{\pi}{2} \:\} \: \: \: \: \: \: \\ \{ \sf \because \: sinx = 0 \implies \: x = n\pi \: where \: n \in \: Z \}$

$\rm \implies\:x = (2n + 1)\dfrac{\pi}{5} \: \: \: or \:x = 2n\pi \: \: where \: n \in \: Z$

Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf T-eq & \bf Solution \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf sinx = 0 & \sf x = n\pi \\ \\ \sf cosx = 0 & \sf x = (2n + 1)\dfrac{\pi}{2}\\ \\ \sf tanx = 0 & \sf x = n\pi\\ \\ \sf sinx = siny & \sf x = n\pi + {( - 1)}^{n}y \\ \\ \sf cosx = cosy & \sf x = 2n\pi \pm \: y\\ \\ \sf tanx = tany & \sf x = n\pi + y \end{array}} \\ \end{gathered}\end{gathered}$

$\: \: \rm \: \: \: \: \: \: \: \: \: \: \: \bull \: \: \: \: \: \: where \: n \in \: Z$