Question

Substitution and Elimination Theta Question Maths​

Answer 1

Step-by-step explanation:

Try by using log x substitute by t

Answer 2

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

Given Trigonometric equation is

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

can be rewritten as

$\rm :\longmapsto\:(sin5\theta + sin\theta ) - sin3\theta= 0$

We know,

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

So, using this identity, we get

$\rm :\longmapsto\:2sin\bigg[\dfrac{5\theta + \theta }{2} \bigg]cos\bigg[\dfrac{5\theta - \theta }{2} \bigg] - sin3\theta = 0$

$\rm :\longmapsto\:2sin3\theta cos2\theta - sin3\theta = 0$

$\rm :\longmapsto\:sin3\theta (2cos2\theta - 1) = 0$

$\rm\implies \:sin3\theta = 0 \: \: or \: \: cos2\theta = \dfrac{1}{2}$

$\rm\implies \:3\theta = \dfrac{\pi}{2} \: \: or \: \: 2\theta = \dfrac{\pi}{3}$

$\rm\implies \:\theta = \dfrac{\pi}{6} \: \: or \: \: \theta = \dfrac{\pi}{6}$

$\bf\implies \:\theta = \dfrac{\pi}{6}$

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Explore More

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