for what values of x where 0< or = x < or = 2pi...?

Question

does sin 4x = -sin 2x.

could somebody show me how to do this not just give me the answer. thanks

Answer 1

sin4x = 2 sin2x cos2x = - sin2x

so solutions are
(a) sin2x = 0 ; when 2x = npi; x = n pi/2 for any integer n
so x = 0, pi/2, pi, 3pi/2, 2pi

(b) cos2x = -1/2; when 2x = 2 pi/3 + 2n pi or 4pi/3 +2n pi
so x = pi/3 + n pi or 2pi/3 + npi
so x = pi/3 or 4pi/3 or 2pi/3 or 5pi/3

overall x= 0, pi/3, pi/2, 2pi/3, pi, 4pi/3, 3pi/2, 5pi/3

Answer 2

If x = 0, then sin 4x = -sin 2x = 0
If x = pi, then sin 4(pi) = 0 = -sin 2(pi) = 0
If x = pi/4, then sin 4(pi/4) = sin 0 =0 and -sin 2(pi/4) not equal o
Now just keep doing this to find the values that make the two functions equal

Answer 3

From the double angle formula...
2 sin 2x cos 2x = -sin 2x
→ cos 2x = -1/2.
→ 2x = (2pi/3 or 4pi/3) + 2*k*pi
→ x = pi/3 or 2pi/3 + k*pi

Solutions: pi/3, 2pi/3, 4pi/3, 5pi/3.

Answer 4

sin 4x = 2 sin 2x * cos 2x

- sin 2x = 2 sin 2x * cos 2x

sin 2x ( 2 cos 2x + 1) = 0

Either sin 2x = 0, i.e 2x = n*PI for every natural number n

x = n*PI/2 n=0,1,2,3,4

Or 2 cos 2x + 1 = 0

2 cos 2x = -1

cos 2x = -0.5

2x = -2pi/3 + 2n*pi or 2x=2pi/3 + 2n*pi

x=-pi/3 + n*pi n=1,2
or
x=pi/3 + n*pi n=0,1