# Angle Question

• Jan 2nd 2006, 04:02 PM
CrimesofParis
Angle 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 PM
ThePerfectHacker
Quote:

Originally Posted by CrimesofParis
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.

Okay you did everything correctly up till now. Now you are required to solved for $x$.
The first one $2\cos x=0$. Divide by two,
$\cos x=0$, now you need to know when $\cos x$ is zero. That is when $x=90,270$ as two non-coterminal angels. To find all solutions it is the same as $x=90+360k$ and $x=270+360k=-90+360+360k=-90+360(1+k)$ where $k$ is any integer. Because as you know that addition and subtraction of 360 keeps the trigonometric function of that angel the same. But $k+1$ is also an integer. Thus, all solutions for $x$ have form. $x=(-1)^{k+1}90+360k, k \epsilon Z$ or in radian form $x=(-1)^{k+1}\frac{\pi}{2}+2\pi k,k \epsilon Z$
• Jan 2nd 2006, 05:47 PM
ThePerfectHacker
$\cos 2x-\sin x=0$
Use the identity
$1-2\sin^2x=\cos 2x$
Thus,
$1-2\sin^2x-\sin x=0$
Change signs,
$2\sin^2x+\sin x-1=0$
Factor,
$(2\sin x-1)(\sin x +1)=0$
Now each one is set to zero,
$2\sin x-1=0$
Thus,
$\sin x=\frac{1}{2}$
That is when $x=30,150$ as the only two non-coterminal angels. Thus all solution have form $x=30+360k, 150+360k$ But,
$x=30+360k=30+180(2k)$
$x=150+360k=-30+180(2k+1)$
Thus, all solutions have form $x=(-1)^k30+180k$ or in radian form $x=(-1)^k\frac{\pi}{6}+\pi k,k \epsilon Z$
• Jan 2nd 2006, 05:55 PM
ThePerfectHacker
Finally the last factor equal to zero, (of the quadratic)
$\sin x+1=0$
Thus,
$\sin x=-1$ That happens when $x=270$ and that is the only non-coterminal angle. Thus all solutions have form $x=270+360k$. But $270+360k=-90+360+360k=-90+360(k+1)$.
Thus, all solutions are $x=-90+360k$. Or in radian form $x=-\frac{\pi}{2}+2\pi k,k\epsilon Z$.

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 PM
CrimesofParis
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.