Question

find general solutions for sin3θ+sin5θ=sin8θ​

Answer 1

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

Given Trigonometric equation is

$\rm :\longmapsto\:sin3\theta + sin5\theta = sin8\theta$

can be rewritten as

$\rm :\longmapsto\:sin5\theta + sin3\theta - sin8\theta = 0$

We know,

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

and

$\boxed{\tt{ sin2x \: = \: 2 \: sinx \: cosx \: }}$

So, using this identity, we get

$\rm :\longmapsto\:2sin\bigg[\dfrac{5\theta + 3\theta }{2} \bigg]cos\bigg[\dfrac{5\theta - 3\theta }{2} \bigg] - 2sin4\theta cos4\theta = 0$

$\rm :\longmapsto\:2sin\bigg[\dfrac{8\theta }{2} \bigg]cos\bigg[\dfrac{2\theta}{2} \bigg] - 2sin4\theta cos4\theta = 0$

$\rm :\longmapsto\:2sin4\theta \: cos\theta \: - \: 2sin4\theta \: cos4\theta = 0$

$\rm :\longmapsto\:2sin4\theta (cos\theta - cos4\theta ) = 0$

We know,

$\boxed{\tt{ cosx - cosy = 2sin\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{y - x}{2} \bigg]}}$

So, using this identity, we get

$\rm :\longmapsto\:2sin4\theta \: \bigg(2sin\bigg[\dfrac{\theta + 4\theta }{2} \bigg]sin\bigg[\dfrac{4\theta - \theta }{2} \bigg] \bigg) = 0$

$\rm :\longmapsto\:4sin4\theta \: sin\bigg[\dfrac{5\theta }{2} \bigg]sin\bigg[\dfrac{3\theta}{2}\bigg] = 0$

We know,

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

So, using this result, we get

$\boxed{\tt{ \sf sin4\theta = 0 \: \: \rm\implies \: \sf 4\theta = n\pi\rm\implies \:\theta = \frac{n\pi}{4} \: \forall \: n \in \: Z}}$

OR

$\boxed{\tt{ \sf sin\dfrac{5\theta }{2} = 0 \: \: \rm\implies \: \sf \dfrac{5\theta }{2} = n\pi\rm\implies \:\theta = \frac{2n\pi}{5} \: \forall \: n \in \: Z}}$

OR

$\boxed{\tt{ \sf sin\dfrac{3\theta }{2} = 0 \: \: \rm\implies \: \sf \dfrac{3\theta }{2} = n\pi\rm\implies \:\theta = \frac{2n\pi}{3} \: \forall \: n \in \: Z}}$

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ADDITIONAL INFORMATION

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