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 March 4th, 2011, 07:50 PM #1 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 Trigonometry $Solve\,for\,x. (1)\,\sin{2x}\,\cos{x}\,-\,2\cos{2x}\,-\,2\sin^3{x}\,=\,0 (2)\,\cos{2x}\,=\,\cos{x}\,+\,\sin{x},\,where\,0\, <\,x\,<\,2\pi$
 March 4th, 2011, 08:51 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Trigonometry 1) $\sin(2x)\cos(x)-2\cos(2x)-2\sin^3(x)=0$ Apply the double-angle identities for sine and cosine: $2\sin(x)\cos^2(x)-2$$\cos^2(x)-\sin^2(x)$$-2\sin^3(x)=0$ Divide through by 2: $\sin(x)\cos^2(x)-\cos^2(x)+\sin^2(x)-\sin^3(x)=0$ Factor by grouping: $\cos^2(x)$$\sin(x)-1$$-\sin^2(x)$$\sin(x)-1$$=0$ $$$\cos^2(x)-\sin^2(x)$$$$\sin(x)-1$$=0$ $\cos(2x)$$\sin(x)-1$$=0$ $x=\frac{\pi}{4},\frac{\pi}{2},\frac{3\pi}{4},\frac {5\pi}{4},\frac{7\pi}{4}$ 2) $\cos(2x)=\cos(x)+\sin(x)$ $\cos^2(x)-\sin^2(x)=\cos(x)+\sin(x)$ $$$\cos(x)+\sin(x)$$$$\cos(x)-\sin(x)$$-$$\cos(x)+\sin(x)$$=0$ $$$\cos(x)+\sin(x)$$$$\cos(x)-\sin(x)-1$$=0$ Set first factor equal to zero: $\cos(x)+\sin(x)=0$ Apply linear combination identity: $\sqrt{2}\sin$$x+\frac{\pi}{4}$$=0$ Setting $x+\frac{\pi}{4}=n\pi$ yields for the interval given: $x=\frac{3\pi}{4},\frac{7\pi}{4}$ Set second factor equal to zero: $\cos(x)-\sin(x)-1=0$ Apply linear combination identity: $\sin$$x+\frac{3\pi}{4}$$=\frac{1}{\sqrt{2}}$ Setting $x+\frac{3\pi}{4}=\frac{pi}{4}+2n\pi$ and $x+\frac{3\pi}{4}=\frac{3pi}{4}+2n\pi$ yields for the given interval: $x=\frac{3\pi}{2}$
March 4th, 2011, 09:01 PM   #3
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Quote:
 Originally Posted by MarkFL $x=\frac{3\pi}{2}$
Correct.
Quote:
 Originally Posted by MarkFL $x=\frac{\pi}{4},\frac{\pi}{2},\frac{3\pi}{4},\frac {5\pi}{4},\frac{7\pi}{4}$
Could you generalize this, where n is any integer?

 March 4th, 2011, 09:18 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Trigonometry For problem 2, I should have put all the solutions together: $x=\frac{3\pi}{4},\frac{3\pi}{2},\frac{7\pi}{4}$ For problem 1, with no restriction on x, we could write the solutions for the two factors as: $x=\frac{(2n+1)\pi}{4},\frac{(4n+1)\pi}{2}$
March 4th, 2011, 09:32 PM   #5
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Quote:
 Originally Posted by MarkFL $x=\frac{(2n+1)\pi}{4},\frac{(4n+1)\pi}{2}$
$x\,=\,n\pi\,\pm\,\frac{\pi}{4}\,\neq\,\frac{(2n\,+ \,1)\pi}{4}$

 March 7th, 2011, 04:04 AM #6 Global Moderator   Joined: Dec 2006 Posts: 20,654 Thanks: 2087 In that case, your textbook's answer is incorrect, since it omits some solutions.
 March 7th, 2011, 01:06 PM #7 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 I knew you were smarter than my textbook's solutions! Thanks for letting me know!

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