solve for x: sin (2x-(pie/6)) - cos (2x-(pie/6)) = 0
0 ≤ x ≤ 2pie
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solve for x: sin (2x-(pie/6)) - cos (2x-(pie/6)) = 0
0 ≤ x ≤ 2pie
Here is one way.Quote:
Originally Posted by rachael
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.
The similar idea here is used as I posted in your other thread.Quote:
Originally Posted by rachael
Ifthen all solutions have form,
But first do what ticbol says to do divide by cosine.
Thus,
Thus, all solutions have form,
Thus,
But,
Thus,
Thus, (solving the inequality)
.
Thus,
because
is an integer.
Thus,
Each one for each value of
Q.E.D.
thank you