Question

Solve for x

$sinx - sin3x + sin5x = 0$

Find general solution​

Answer 1

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

Given Trigonometric equation is

$\rm :\longmapsto\: \: sinx - sin3x + sin5x = 0$

can be rewritten as

$\rm :\longmapsto\:(sin5x + sinx) - sin3x = 0$

We know,

$\red{\boxed{\tt{ sinx + siny = 2sin\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg]}}}$

So, using this identity, we get

$\rm :\longmapsto\:2sin\bigg[\dfrac{5x + x}{2} \bigg]cos\bigg[\dfrac{5x - x}{2} \bigg] - sin3x = 0$

$\rm :\longmapsto\:2sin3x \: cos2x - sin3x= 0$

$\rm :\longmapsto\:sin3x(2cos2x - 1) = 0$

$\rm\implies \:sin3x = 0 \: \: \: or \: \: \: cos2x = \dfrac{1}{2}$

Now, Consider

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

We know,

$\red{\boxed{\tt{ sinx = 0 \rm\implies \:\sf x = n\pi \: \forall \: n \in \: Z}}}$

$\rm\implies \:3x = n\pi \: \forall \: n \in \: Z$

$\rm\implies \:\boxed{\tt{ x = \dfrac{n\pi}{3} \: \forall \: n \in \: Z}}$

Now, Consider

$\rm :\longmapsto\:cos2x = \dfrac{1}{2}$

$\rm :\longmapsto\:cos2x =cos \dfrac{\pi}{3}$

We know,

$\red{\boxed{\tt{ cosx = cosy \: \rm\implies \: \sf x = 2n\pi \pm \: y\: \forall \: n \in \: Z}}}$

So, using this identity, we get

$\rm \longmapsto\: 2x = 2n\pi \pm \: \dfrac{\pi}{3} \: \forall \: n \in \: Z$

$\rm\implies \: \boxed{\tt{ \: x = n\pi \pm \: \dfrac{\pi}{6} \: \forall \: n \in \: Z \: }}$

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More to Know

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

Answer 2

Answer:

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

Given Trigonometric equation is

$\rm :\longmapsto\: \: sinx - sin3x + sin5x = 0$

can be rewritten as

$\rm :\longmapsto\:(sin5x + sinx) - sin3x = 0$

We know,

$\red{\boxed{\tt{ sinx + siny = 2sin\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg]}}}$

So, using this identity, we get

$\rm :\longmapsto\:2sin\bigg[\dfrac{5x + x}{2} \bigg]cos\bigg[\dfrac{5x - x}{2} \bigg] - sin3x = 0$

$\rm :\longmapsto\:2sin3x \: cos2x - sin3x= 0$

$\rm :\longmapsto\:sin3x(2cos2x - 1) = 0$

$\rm\implies \:sin3x = 0 \: \: \: or \: \: \: cos2x = \dfrac{1}{2}$

Now, Consider

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

We know,

$\red{\boxed{\tt{ sinx = 0 \rm\implies \:\sf x = n\pi \: \forall \: n \in \: Z}}}$

$\rm\implies \:3x = n\pi \: \forall \: n \in \: Z$

$\rm\implies \:\boxed{\tt{ x = \dfrac{n\pi}{3} \: \forall \: n \in \: Z}}$

Now, Consider

$\rm :\longmapsto\:cos2x = \dfrac{1}{2}$

$\rm :\longmapsto\:cos2x =cos \dfrac{\pi}{3}$

We know,

$\red{\boxed{\tt{ cosx = cosy \: \rm\implies \: \sf x = 2n\pi \pm \: y\: \forall \: n \in \: Z}}}$

So, using this identity, we get

$\rm \longmapsto\: 2x = 2n\pi \pm \: \dfrac{\pi}{3} \: \forall \: n \in \: Z$

$\rm\implies \: \boxed{\tt{ \: x = n\pi \pm \: \dfrac{\pi}{6} \: \forall \: n \in \: Z \: }}$

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