solve for x: sin (2x-(pie/6)) - cos (2x-(pie/6)) = 0

0 ≤ x ≤ 2pie

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- February 19th 2006, 11:43 PMrachaeltri
solve for x: sin (2x-(pie/6)) - cos (2x-(pie/6)) = 0

0 ≤ x ≤ 2pie - February 20th 2006, 02:42 AMticbolQuote:

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. - February 20th 2006, 11:28 AMThePerfectHackerQuote:

Originally Posted by**rachael**

If then 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. - February 20th 2006, 10:11 PMrachael
thank you