i need to clarify something about trigo?

Question

can anyone please tell me how the x is found in details? thanks in advance.

i) sin 4x = 0 (0 to 2pi interval)
x = 0, pi/4, pi/2, 3pi/4, 1pi

ii) cos (-2x) = 1 (0 to 2pi interval)
x = 0, 1pi

Answer 1

sin 4x = 0
4x = inv Sin 0
Keeping in mind that the solution i get for the above is only one answer. I will generate a complete set of solutions for 4x, then divide by 4
4x = 0 + n (2pi) (all solutions coterminal with 0, where n is an integer.)
plus
4x = pi + n(2pi) this is all solutions coterminal with pi, another value where the sine is 0)
Dividing by 4 we get
x = 0 + n(pi/2) and x = pi/4 + n(pi/2)
Since n is an integer and the required domain is 0 to 2pi all you need to do to get all the solutions is substitute integer values for n that will give you solutions in the required domain.
x = 0, pi/4, pi/2, 3pi/4, pi, 5pi/4, 3pi/2, 7pi/4

Now consider cos(-2x) = 1
-2x = inv Cos (1)
-2x = pi/2 + n (2pi)
x = - pi - n(pi)
Keeping in mind that n is an integer I will substitute negatives as necessary to get answers in the required domain.
x = 0, pi
Hope this helps.

Answer 2

You need to know all the special angles and their trig values:
memorise 0, 30, 45, 60, and 90 degrees (know their radian equivalents too) their sin, cos, tan because you will need them often as you do here.
sin of 0 and 180 (0 and pi radians) and their multiples are all equal to zero.
if the interval of x is 0 to 2i then:
sin 0 = 0 and sin pi = 0 sin 2pi = 0, sin 3pi =0, sin 4pi = 0, sin 5pi= 0, sin 6pi = 0, sin 7pi = 0, sin 8pi = 0
SO:
4x= 0, 4x=pi, 4x = 2pi, 4x = 3pi, 4x = 4pi, 4x = 5pi, 4x = 6pi, 4x = 7pi, 4x = 8pi.
x=0, x=pi/4, x= pi/2, x= 3/4 pi, x=pi, x= 5/4 pi, x=6/4 pi, x=7/4 pi, x=8/4 pi = 2pi.

for the second one the cos of 0 is 1 then the values at which (-2x) = 0 are:
cos 0 = 0 and cos -pi = 0 cos -2pi = 0, cos -3pi =0, cos -4pi = 0

-2x= 0, -2x= -pi, -2x = -2pi, -2x = -3pi, -2x = -4pi SO:
x = 0, x = pi/2, x= pi, x=2/3 pi, x= 2pi.

Answer 3

i) sin4x = 0 = sin 0
=> 4x = k(pi) + (- 1)^k (0)
=> x = k (pi/4), k belongs to R
Taking k = 0, 1, 2, 3, 4, 5, 6, 7, 8 (assuming you mean interval [ 0, 2(pi) ])
=> x = 0, pi/4, pi/2, 3pi/4, pi, 5pi/4, 3pi/2, 7pi/4, 2pi

ii) cos ( - 2x) = 1 = cos 0
=> - 2x = 2k(pi) +/- 0
=> x = - k (pi), k belongs to R
Taking k = 0, -1, -2
=> x = 0, pi, 2pi