So I have two angle problems...

sin2x = 2cosxcos2x ; which becomes

0 = 2cosxcos2x - sin2x

0 = 2cosxcos2x - 2sinxcosx

0 = 2cosx(cos2x - sin x)

so either

2cosx = 0 OR cos2x - sinx = 0

Thank you.

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- Jan 2nd 2006, 04:02 PMCrimesofParisAngle Question
So I have two angle problems...

sin2x = 2cosxcos2x ; which becomes

0 = 2cosxcos2x - sin2x

0 = 2cosxcos2x - 2sinxcosx

0 = 2cosx(cos2x - sin x)

so either

2cosx = 0 OR cos2x - sinx = 0

Thank you. - Jan 2nd 2006, 05:30 PMThePerfectHackerQuote:

Originally Posted by**CrimesofParis**

Okay you did everything correctly up till now. Now you are required to solved for $\displaystyle x$.

The first one $\displaystyle 2\cos x=0$. Divide by two,

$\displaystyle \cos x=0$, now you need to know when $\displaystyle \cos x$ is zero. That is when $\displaystyle x=90,270$ as two non-coterminal angels. To find all solutions it is the same as $\displaystyle x=90+360k$ and $\displaystyle x=270+360k=-90+360+360k=-90+360(1+k)$ where $\displaystyle k$ is any integer. Because as you know that addition and subtraction of 360 keeps the trigonometric function of that angel the same. But $\displaystyle k+1$ is also an integer. Thus, all solutions for $\displaystyle x$ have form. $\displaystyle x=(-1)^{k+1}90+360k, k \epsilon Z $ or in radian form $\displaystyle x=(-1)^{k+1}\frac{\pi}{2}+2\pi k,k \epsilon Z$ - Jan 2nd 2006, 05:47 PMThePerfectHacker
For your second question:

$\displaystyle \cos 2x-\sin x=0$

Use the identity

$\displaystyle 1-2\sin^2x=\cos 2x$

Thus,

$\displaystyle 1-2\sin^2x-\sin x=0$

Change signs,

$\displaystyle 2\sin^2x+\sin x-1=0$

Factor,

$\displaystyle (2\sin x-1)(\sin x +1)=0$

Now each one is set to zero,

$\displaystyle 2\sin x-1=0$

Thus,

$\displaystyle \sin x=\frac{1}{2}$

That is when $\displaystyle x=30,150$ as the only two non-coterminal angels. Thus all solution have form $\displaystyle x=30+360k, 150+360k$ But,

$\displaystyle x=30+360k=30+180(2k)$

$\displaystyle x=150+360k=-30+180(2k+1)$

Thus, all solutions have form $\displaystyle x=(-1)^k30+180k$ or in radian form $\displaystyle x=(-1)^k\frac{\pi}{6}+\pi k,k \epsilon Z$ - Jan 2nd 2006, 05:55 PMThePerfectHacker
Finally the last factor equal to zero, (of the quadratic)

$\displaystyle \sin x+1=0$

Thus,

$\displaystyle \sin x=-1$ That happens when $\displaystyle x=270$ and that is the only non-coterminal angle. Thus all solutions have form $\displaystyle x=270+360k$. But $\displaystyle 270+360k=-90+360+360k=-90+360(k+1)$.

Thus, all solutions are $\displaystyle x=-90+360k$. Or in radian form $\displaystyle x=-\frac{\pi}{2}+2\pi k,k\epsilon Z$.

If you have difficulty understanding all that manipulating after the solutions just understand that by adding 360 or subtracting 360 you change nothing. Thus, those are**all**the solutions. All I did was manipulated into a more elegant form. - Jan 2nd 2006, 06:36 PMCrimesofParis
Thank you so much, that pretty much answers my questions, except I forgot to mention the answer has to be radians, so I should be able to covert the middle steps to radians. But thank you.