Question

Solve the equation:
➜Sin 2x = sinx, -Π ≤ Π​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Appropriate Question :-

Solve the equation :-

$\rm \: sin2x = sinx,\: \: - \pi \leqslant x \leqslant \pi$

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

Given Trigonometric equation is

$\rm \: sin2x = sinx\:$

$\rm \: sin2x - sinx = 0\:$

$\rm \: 2 \: sinx \: cosx - sinx = 0\:$

$\rm \: sinx \: (2cosx - 1) = 0$

$\rm \: sinx = 0 \: \: \: or \: \: \: cosx = \dfrac{1}{2}$

Case :- 1

$\rm \: sinx = 0$

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

As it is given that,

$\rm \: - \pi \leqslant x \leqslant \pi$

$\rm\implies \:x = - \pi, \: 0, \: \pi$

Case :- 2

$\rm \:\: cosx = \dfrac{1}{2}$

$\rm \:\: cosx = cos \dfrac{\pi}{3}$

We know,

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

So, using this, we get

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

As it is given that

$\rm \: - \pi \leqslant x \leqslant \pi$

$\rm\implies \:x = - \dfrac{2\pi}{3}, \: - \dfrac{\pi}{3}, \: \dfrac{\pi}{3}, \: \dfrac{2\pi}{3}$

Hence, Solution is

$\rm\implies \:x = - \pi, \: - \dfrac{2\pi}{3}, \: - \dfrac{\pi}{3}, \:0, \: \dfrac{\pi}{3}, \: \dfrac{2\pi}{3}, \: \pi$

$\rule{190pt}{2pt}$

Additional Information :-

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