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Question Number 17153 by Tinkutara last updated on 01/Jul/17

If m, n ∈ N(n > m), then number of  solutions of the equation  n∣sin x∣ = m∣sin x∣ in [0, 2π] is

$$\mathrm{If}\:{m},\:{n}\:\in\:{N}\left({n}\:>\:{m}\right),\:\mathrm{then}\:\mathrm{number}\:\mathrm{of} \\ $$ $$\mathrm{solutions}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$ $${n}\mid\mathrm{sin}\:{x}\mid\:=\:{m}\mid\mathrm{sin}\:{x}\mid\:\mathrm{in}\:\left[\mathrm{0},\:\mathrm{2}\pi\right]\:\mathrm{is} \\ $$

Answered by mrW1 last updated on 01/Jul/17

n∣sin x∣ = m∣sin x∣  ∵ m≠n  ∴ ∣sin x∣=0  x=0,π,2π  ⇒there are 3 solutions

$${n}\mid\mathrm{sin}\:{x}\mid\:=\:{m}\mid\mathrm{sin}\:{x}\mid \\ $$ $$\because\:\mathrm{m}\neq\mathrm{n} \\ $$ $$\therefore\:\mid\mathrm{sin}\:\mathrm{x}\mid=\mathrm{0} \\ $$ $$\mathrm{x}=\mathrm{0},\pi,\mathrm{2}\pi \\ $$ $$\Rightarrow\mathrm{there}\:\mathrm{are}\:\mathrm{3}\:\mathrm{solutions} \\ $$

Commented byTinkutara last updated on 02/Jul/17

Thanks Sir!

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

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