sin-2x-pi-4-sin-2x-pi-3-0-x-

Question Number 132347 by Study last updated on 13/Feb/21

s i n (2 x + \frac{π}{4}) - s i n (2 x + \frac{π}{3}) = 0 x = ?

Answered by EDWIN88 last updated on 13/Feb/21

\sin X - \sin Y = 2 \cos (\frac{X + Y}{2}) \sin (\frac{X - Y}{2})

where {\begin{cases} X = 2 x + \frac{π}{4} \\ Y = 2 x + \frac{π}{3} \end{cases}

\sin (2 x + \frac{π}{4}) - \sin (2 x + \frac{π}{3}) = 0

2 \cos (2 x + \frac{7 π}{24}) . \sin (- \frac{π}{24}) = 0

\cos (2 x + \frac{7 π}{24}) = 0 = \cos \frac{π}{2}

2 x = \pm \frac{π}{2} - \frac{7 π}{24} + 2 n π

x = \pm \frac{π}{4} - \frac{7 π}{48} + n π \Rightarrow x = {\begin{cases} \frac{5 π}{48} + n π \\ - \frac{19}{48} + n π \end{cases}

Answered by Ar Brandon last updated on 13/Feb/21

\sin (2 x + \frac{π}{4}) - \sin (2 x + \frac{π}{3}) = 0 \Rightarrow \sin (2 x + \frac{π}{4}) = \sin (2 x + \frac{π}{3})

{\begin{cases} 2 x + \frac{π}{4} = 2 x + \frac{π}{3} \\ 2 x + \frac{π}{4} = π - (2 x + \frac{π}{3}) \end{cases} \Rightarrow 4 x = π - \frac{π}{4} - \frac{π}{3} = \frac{5 π}{12} [2 π]

\Rightarrow x = \frac{5 π}{48} [\frac{π}{2}]