# Solve trig equation

• Apr 18th 2006, 08:16 AM
ling_c_0202
Solve trig equation
I can't get the correct ans for this question~~

Solve the equation for 0 deg <= theta <= 360 deg.

$\displaystyle \sin 5 \theta/ \sin \theta - \ cos 5 \theta/\cos \theta = 2$
• Apr 18th 2006, 09:22 AM
ThePerfectHacker
Quote:

Originally Posted by ling_c_0202
I can't get the correct ans for this question~~

Solve the equation for 0 deg <= theta <= 360 deg.

$\displaystyle \sin 5 \theta/ \sin \theta - \ cos 5 \theta/\cos \theta = 2$

Add fractions (remember commom denominator :eek: ):
$\displaystyle \frac{\sin 5x \cos x-\sin x\cos 5x}{\sin x\cos x}=2$
But the numerator is expansion of $\displaystyle \sin (5x-x)=\sin 4x$
Thus, you have,
$\displaystyle \frac{\sin 4x}{\sin x\cos x}=2$
Multiply both sides by a half to get,
$\displaystyle \frac{\sin 4x}{2 \sin x\cos x}=1$
But, $\displaystyle 2\sin x\cos x=\sin 2x$ and $\displaystyle \sin 4x=2\sin 2x\cos 2x$ (double angles),
$\displaystyle \frac{2\sin 2x\cos 2x}{\sin 2x}=1$
Notice the $\displaystyle \sin 2x$ kills the other $\displaystyle \sin 2x$ thus,
$\displaystyle 2\cos 2x=1$
Thus,
$\displaystyle \cos 2x=1/2$
Thus, all the solutions take the form,
$\displaystyle 2x=\left\{ \begin{array} {c}\frac{\pi}{3}+2\pi k\\ -\frac{\pi}{3}+2\pi k$
Thus,
$\displaystyle x=\left\{ \begin{array}{c} \frac{\pi}{6}+\pi k\\ -\frac{\pi}{6}+\pi k$
You need them to be such as, $\displaystyle 0\leq x\leq 2\pi$
For the example the top solution is true only for $\displaystyle k=1$ otherwise it gets too big or too small. Substituting $\displaystyle k=1$ for the top solution we get,
$\displaystyle \frac{\pi}{6}+\pi=\frac{7\pi}{6}$
The bottom solution is true only for $\displaystyle k=1,2$ otherwise it gets too big or too small. Substituting $\displaystyle k=1,2$ for the bottom solution we get,
$\displaystyle -\frac{\pi}{6}+\pi=\frac{5\pi}{6}$
$\displaystyle -\frac{\pi}{6}+2\pi=\frac{11\pi}{6}$
Thus,
$\displaystyle x=\frac{5\pi}{6},\frac{7\pi}{6},\frac{11\pi}{6}$
• Apr 19th 2006, 08:23 AM
ling_c_0202
but for the step
$\displaystyle \frac{2\sin 2x\cos 2x}{\sin 2x}=1$

it is not given in the question sin 2x <> 0,

how come they could kill each other?
:confused:
is it possible if i 'lost' any solution?
• Apr 19th 2006, 10:41 AM
ThePerfectHacker
Quote:

Originally Posted by ling_c_0202
but for the step
$\displaystyle \frac{2\sin 2x\cos 2x}{\sin 2x}=1$

it is not given in the question sin 2x <> 0,

how come they could kill each other?
:confused:
is it possible if i 'lost' any solution?

Because $\displaystyle \frac{\sin 2x}{\sin 2x}=1$ for defined values of $\displaystyle x$
• Apr 20th 2006, 06:28 AM
ling_c_0202
:p :D THANK YOU!! :)