Trigonometric equation

man111 · 21-11-2010, 08:27 PM

$Q:$\Rightarrow$ Show that the Trigonometric equation $\sin x+2\sin 2x=3+\sin 3x$ has no Roots $x\in [0,2\pi].$
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Show that the equation has no real roots..

toandhsp · 21-11-2010, 10:01 PM

Chú ý
$\sin x + 2\sin 2x = 3 + \sin 3x $ tương đương với
$(\sin x + \cos 2x)^2 + (\sin 2x - 1)^2 + \cos^2 x = 0. $

man111 · 22-11-2010, 01:37 PM

Thanks Toandshop.
$$\boldsymbol{Ans:}$\Rightarrow$ Here $sinx+sin2x=3+sin3x\Leftrightarrow3+sin3x=sinx+sin2x$\\\\ $3+\underbrace{sin3x-sinx}-2sin2x=0\Leftrightarrow3+2cos2xsinx-2sin2x=0$\\\\ $\underbrace{cos^22x+sin^22x}_{1}+\underbrace{cos^2x+sin^2x}_{1}+\underbrace{1}+2cos2x.sinx-2sin2x=0$\\\\ Now Rearranging We get\\\\ $\underbrace{cos^22x+sin^2x+2cos2x.sinx}+\underbrace{1+sin^22x-2sin2x}+cos^2x=0$\\\\ $\underbrace{(cos2x+sinx)^2}+\underbrace{(sin2x-1)^2}+(cosx)^2=0$....................................(1)\\\\ Which is possible only When $cos2x+sinx=0$ and $sin2x-1=0$ and $cosx=0$\\\\ solve $cos2x+sinx=0\Leftrightarrow1-2sin^2x+sinx=0$,We get $\boxed{x=\frac{\pi}{2},\frac{7\pi}{6},\frac{11\pi}{6}}$\\\\ similarly $sin2x-1=0$ We Get $\boxed{x=\frac{\pi}{4},\frac{5\pi}{4}}$\\\\ $cosx=0$,We Get $\boxed{x=\frac{\pi}{2},\frac{3\pi}{2}}$,Here $x\epsilon[0,2\pi]$\\\\ So from 3-equation , No Common value of $x$. So No solution of This Trigonometric equation.$

21-11-2010, 08:27 PM	#1
man111 +Thành Viên+ Tham gia ngày: Nov 2010 Bài gởi: 280 Thanks: 152 Thanked 77 Times in 49 Posts	Trigonometric equation $Q:$\Rightarrow$ Show that the Trigonometric equation $\sin x+2\sin 2x=3+\sin 3x$ has no Roots $x\in [0,2\pi].$ ------------------------------ Show that the equation has no real roots.. thay đổi nội dung bởi: novae, 21-11-2010 lúc 09:18 PM Lý do: Tự động gộp bài

21-11-2010, 10:01 PM	#2
toandhsp +Thành Viên+ Tham gia ngày: Jul 2008 Bài gởi: 27 Thanks: 5 Thanked 24 Times in 11 Posts	Chú ý $\sin x + 2\sin 2x = 3 + \sin 3x $ tương đương với $(\sin x + \cos 2x)^2 + (\sin 2x - 1)^2 + \cos^2 x = 0. $

22-11-2010, 01:37 PM	#3
man111 +Thành Viên+ Tham gia ngày: Nov 2010 Bài gởi: 280 Thanks: 152 Thanked 77 Times in 49 Posts	Thanks Toandshop. $$\boldsymbol{Ans:}$\Rightarrow$ Here $sinx+sin2x=3+sin3x\Leftrightarrow3+sin3x=sinx+sin2x$\\\\ $3+\underbrace{sin3x-sinx}-2sin2x=0\Leftrightarrow3+2cos2xsinx-2sin2x=0$\\\\ $\underbrace{cos^22x+sin^22x}_{1}+\underbrace{cos^2x+sin^2x}_{1}+\underbrace{1}+2cos2x.sinx-2sin2x=0$\\\\ Now Rearranging We get\\\\ $\underbrace{cos^22x+sin^2x+2cos2x.sinx}+\underbrace{1+sin^22x-2sin2x}+cos^2x=0$\\\\ $\underbrace{(cos2x+sinx)^2}+\underbrace{(sin2x-1)^2}+(cosx)^2=0$....................................(1)\\\\ Which is possible only When $cos2x+sinx=0$ and $sin2x-1=0$ and $cosx=0$\\\\ solve $cos2x+sinx=0\Leftrightarrow1-2sin^2x+sinx=0$,We get $\boxed{x=\frac{\pi}{2},\frac{7\pi}{6},\frac{11\pi}{6}}$\\\\ similarly $sin2x-1=0$ We Get $\boxed{x=\frac{\pi}{4},\frac{5\pi}{4}}$\\\\ $cosx=0$,We Get $\boxed{x=\frac{\pi}{2},\frac{3\pi}{2}}$,Here $x\epsilon[0,2\pi]$\\\\ So from 3-equation , No Common value of $x$. So No solution of This Trigonometric equation.$