tri

• Feb 19th 2006, 11:43 PM
rachael
tri
solve for x: sin (2x-(pie/6)) - cos (2x-(pie/6)) = 0
0 ≤ x ≤ 2pie
• Feb 20th 2006, 02:42 AM
ticbol
Quote:

Originally Posted by rachael
solve for x: sin (2x-(pie/6)) - cos (2x-(pie/6)) = 0
0 ≤ x ≤ 2pie

Here is one way.

sin(2x -pi/6) -cos(2x -pi/6) = 0
sin(2x -pi/6) = cos(2x -pi/6)
Divide both sides by cos(2x -pi/6),
tan(2x -pi/6) = 1

A positive tangent value. Meaning, angle (2x -pi/6) is in the 1st or 3rd quadrants.

(2x -pi/6) = arctan(1)
(2x -pi/6) = pi/4 ---------------in the 1st quadrant,
or,
(2x -pi/6) = pi +pi/4 = 5pi/4 ----in the 3rd quadrant.

So,
2x -pi/6 = pi/4
2x = pi/4 +pi/6
2x = (6pi +4pi)/(4*6)
2x = 10pi/24
x = 10pi/(24*2) = 5pi/24 = (5/24)pi radians, or 37.5 degrees -----answer.

Or,
2x -pi/6 = 5pi/4
2x = 5pi/4 +pi/6
2x = (6*5pi +4pi)/(4*6)
2x = 34pi/24
x = 34pi/(24*2) = 17pi/24 = (17/24)pi rad, or 127.5 deg -----answer.

-----------------------------
Check, in degrees, for easier use of calculator,

When x = 37.5 degrees,
sin(2x -pi/6) -cos(2x -pi/6) = 0
sin(2*37.5deg -30deg) -cos(2*37.5deg -30deg) =? 0
sin(75deg -30deg) -cos(75deg -30deg) =? 0
sin(45deg) -cos(45deg) =? 0
1/sqrt(2) -1/sqrt(2) =? 0
0 =? 0
Yes, so, OK.

When x = 127.5 degrees,
sin(2x -pi/6) -cos(2x -pi/6) = 0
sin(2*127.5deg -30deg) -cos(2*17.5deg -30deg) =? 0
sin(255deg -30deg) -cos(255deg -30deg) =? 0
sin(225deg) -cos(225deg) =? 0
-0.7071 -(-0.7071) =? 0
0 =? 0
Yes, so, OK.
• Feb 20th 2006, 11:28 AM
ThePerfectHacker
Quote:

Originally Posted by rachael
solve for x: sin (2x-(pie/6)) - cos (2x-(pie/6)) = 0
0 ≤ x ≤ 2pie

The similar idea here is used as I posted in your other thread.
If $\displaystyle \tan x=1$ then all solutions have form,
$\displaystyle x=\frac{\pi}{4}+\pi k$

But first do what ticbol says to do divide by cosine.
Thus,
$\displaystyle \tan\left( 2x-\frac{\pi}{6}\right)=1$
Thus, all solutions have form,
$\displaystyle 2x-\frac{\pi}{6}=\frac{\pi}{4}+\pi k$
Thus,
$\displaystyle x=\frac{5\pi}{24}+\frac{\pi}{2}=12.5^o+90^ok$
But, $\displaystyle 0^o\leq x\leq 360^o$
Thus,
$\displaystyle 0^o\leq12.5^o+90^ok\leq 360$
Thus, (solving the inequality)
$\displaystyle -.13\bar 8\leq k\leq 3.86\bar 1$.
Thus,
$\displaystyle k=0,1,2,3$ because $\displaystyle k$ is an integer.
Thus,
$\displaystyle x=12.5^o$
$\displaystyle x=102.5^o$
$\displaystyle x=192.5^o$
$\displaystyle x=282.5^o$
Each one for each value of $\displaystyle k$.
Q.E.D.
• Feb 20th 2006, 10:11 PM
rachael
thank you