Trigonometry

MarkFL · March 4th, 2011, 08:51 PM

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:

$x=\frac{3\pi}{2}$

johnny · March 4th, 2011, 09:01 PM

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?

MarkFL · March 4th, 2011, 09:18 PM

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:

$x=\frac{(2n+1)\pi}{4},\frac{(4n+1)\pi}{2}$

johnny · March 4th, 2011, 09:32 PM

Quote:

Originally Posted by MarkFL

$x=\frac{(2n+1)\pi}{4},\frac{(4n+1)\pi}{2}$

$x\,=\,n\pi\,\pm\,\frac{\pi}{4}\,\neq\,\frac{(2n\,+ \,1)\pi}{4}$
My textbook's answer.

skipjack · March 7th, 2011, 04:04 AM

In that case, your textbook's answer is incorrect, since it omits some solutions.

johnny · March 7th, 2011, 01:06 PM

I knew you were smarter than my textbook's solutions! Thanks for letting me know!

March 4th, 2011, 07:50 PM	#1
johnny Senior Member Joined: Apr 2007 Posts: 2,140 Thanks: 0	Trigonometry $Solve\,for\,x.<br /> (1)\,\sin{2x}\,\cos{x}\,-\,2\cos{2x}\,-\,2\sin^3{x}\,=\,0<br /> (2)\,\cos{2x}\,=\,\cos{x}\,+\,\sin{x},\,where\,0\, <\,x\,<\,2\pi$
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March 4th, 2011, 08:51 PM	#2
MarkFL 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: $x=\frac{3\pi}{2}$
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March 4th, 2011, 09:18 PM	#4
MarkFL 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: $x=\frac{(2n+1)\pi}{4},\frac{(4n+1)\pi}{2}$
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March 7th, 2011, 04:04 AM	#6
skipjack Global Moderator Joined: Dec 2006 Posts: 20,654 Thanks: 2087	In that case, your textbook's answer is incorrect, since it omits some solutions.
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March 7th, 2011, 01:06 PM	#7
johnny 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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