# Questions on trigo. equation

• Sep 16th 2007, 10:45 PM
acc100jt
Questions on trigo. equation
Find general solution for the following
1. $\cos{3x}=2\cos{x}-\sqrt{3}\sin{3x}$
2. $\cos{(2x+60^o)}=\sin{x}$

can anyone help?
• Sep 17th 2007, 03:35 AM
Jhevon
Quote:

Originally Posted by acc100jt
Find general solution for the following
1. $\cos{3x}=2\cos{x}-\sqrt{3}\sin{3x}$
2. $\cos{(2x+60^o)}=\sin{x}$

can anyone help?

expand using the addition and double angle formulas and then see what you can do.

recall:

$\cos 2 \theta = \cos^2 \theta - \sin^2 \theta = 2 \cos^2 \theta - 1 = 1 - 2 \sin^2 \theta$

$\sin 2 \theta = 2 \sin \theta \cos \theta$

$\cos ( \alpha + \beta ) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$

note: $\cos 3 x = \cos (2x + x)$ and $\sin 3x = \sin (2x + x)$
• Sep 17th 2007, 09:57 AM
red_dog
Quote:

Originally Posted by acc100jt
Find general solution for the following
1. $\cos{3x}=2\cos{x}-\sqrt{3}\sin{3x}$

$\cos 3x+\sqrt{3}\sin 3x=2\cos x$
Divide both members by 2:
$\displaystyle\frac{1}{2}\cos 3x+\frac{\sqrt{3}}{2}\sin 3x=\cos x$
$\cos 60^{\circ}\cos 3x+\sin 60^{\circ}\sin 3x=\cos x$
$\cos(60^{\circ}-3x)=\cos x$
$\cos(60^{\circ}-3x)-\cos x=0$
$\displaystyle 2\sin(2x-30^{\circ})\sin (30^{\circ}-x)=0$

Now, continue.
• Sep 17th 2007, 10:56 AM
ticbol
Quote:

Originally Posted by acc100jt
Find general solution for the following
1. $\cos{3x}=2\cos{x}-\sqrt{3}\sin{3x}$
2. $\cos{(2x+60^o)}=\sin{x}$

can anyone help?

Umm, these are really very hard to solve. Need tricks.

Even when I was yet in High School, I resort to iteration, or trial and error, in solving equations that I didn't know how to solve the usual ways.
Problem 1 is is a good candidate for my iteration.

By luck, or because of the sqrt(3) in the equation, I tried first, x = 30degrees.
cos(3*30deg) =? 2cos(30deg) -sqrt(3)*sin(3*30deg)
cos(90deg) =? 2cos(30deg) -sqrt(3)*sin(90deg)
0 =? 2[sqrt(3) / 2] -sqrt(3)*(1)
0 =? sqrt(3) -sqrt(3)
0 =? 0
Yes, so x = 30deg is OK.

I was just lucky to hit it in one try only. So, 30deg is a solution. There might be some more but I don't know how, or I don't want, to continue after that.

------------------------------
For the second problem, another trick.

cos(2x +60deg) = sin(x) ----------(ii)

Expanding that will lead to an equation that is as complicated as the first one, so another dip into our bag of tricks:

We know that cosA = sin(90deg -A).
That means A +(90deg -A) = 90 deg.

Or, cosA = sinB -----if angles A and B are complementary.

So, using that on (ii),
(2x +60deg) +x = 90deg
3x = 90 -60
x = 30/3
x = 10 deg ------------a solution.

Check,
cos(2*10deg +60deg) =? sin(10deg)
cos(80deg) =? sin(10deg)
Using a calculator,
0.173648178 =? 0.173648178
Yes, so, x = 10 degrees is correct.
• Sep 17th 2007, 07:01 PM
Jhevon
Quote:

Originally Posted by red_dog
$\cos 3x+\sqrt{3}\sin 3x=2\cos x$
Divide both members by 2:
$\displaystyle\frac{1}{2}\cos 3x+\frac{\sqrt{3}}{2}\sin 3x=\cos x$
$\cos 60^{\circ}\cos 3x+\sin 60^{\circ}\sin 3x=\cos x$
$\cos(60^{\circ}-3x)=\cos x$
$\cos(60^{\circ}-3x)-\cos x=0$
$\displaystyle 2\sin(2x-30^{\circ})\sin (30^{\circ}-x)=0$

Now, continue.

nice!