0=sin3x-cos3x ; 0<=x=< 2pi?

Question

determing the intersection of petals for the equations of sin3x and
cos3x

Answer 1

0 = sin(3x) - cos(3x)
cos(3x) = sin(3x)
Divide both sides by cos(3x) to get 1 = tan(3x).

In the range 0 <= x <= 2π, tan(x) = 1 only for π/4 and 5π/4. So that means x = π/12 and 5π/12

Answer 2

cos(3x)=1/2 3x = cos-1(1/2) x = 1/3(cos-1(1/2)) Put that in your calculator and find the value that is between 0 and 2pi, if the first value that comes up i <0 or >2pi then draw out the cos graph and find the point that is >0 and >2pi on the X axis with the same value on the Y axis and use logic to find out the exact point on the X axis (eg if you wanted a number between -90 and 0 and you got 10, you'd subtract 10 to get 80 as it's symmetrical on the y axis at 0 so the point 10 above 0 has the same value as the point 10 below etc.) That is the longest to explain but most logical and accurate way to work it out, my dad who has a PHd and researches/lectures at St Andrews University (best Uni in Scotland and in the top 5 in the UK) in Applied mathematics (The mathematics department is apparently the most prestigious part of the Uni) taught me this way, the way I was taught in school was basically a crappy formulaic thing that required no understanding, only memorising a set of instructions.

Answer 3

0 = sin3x - cos3x
sin3x = cos3x
3x = π/4, 5π/4
x = π/12, 5π/12

Answer 4

0=sin3x-cos3x

cos3x = sin3x

tan3x = 1

3x = π/4, 5π/4, 9π/4, 13π/4, 17π/4 and 21π/4

x = π/12, 5π/12, 9π/12, 13π/12, 17π/12 and 21π/12

Answer 5

sin 3x / cos3x = 1
tan 3x = 1
3x in 1st and 3 rd quadrants
3x = π/4 and π/4 + 2kπ
x = π/12 and π/12 + 2kπ/3 (integer k)

3x = 3π/4 and 3π/4 + 2kπ
x = π/4 and π/4 + 2kπ/3 (integer k)

Answer 6

Let z = 3x

Then sinz - cosz = 0

sinz = cosz

This occurs when z = pi/2 , or 5pi/2

z = pi/2
Then x = pi/6

z = 5pi/2
Then x = 5pi/6

Answer 7

Does one cancel the other out?

Answer 8

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