Find all the solutions and indicate solutions in interval [0,2π)?
sin2x - sinx cosx = cosx
i need to see the work, that would help a lot.
- panic modeLv 78 years agoFavorite Answer
is the first sin term of 2x or it is square of sin x?
- Adrian SLv 78 years ago
Using the double angle identity for sin(2x) you'll get 2sin(x)cos(x)
Now subtract and bring over the cos(x)and set = 0 and you'll get
cos(x)*(sin(x) -1) = 0.
By ZPP cos(x) = 0 when x = pi/2 and 3pi/2
and sin(x) = 1 when x = pi/2Source(s): Math tutor for several years.
- KateLv 48 years ago
sin 2x - sin x cos x = cos x
Substitute in sin2x = 2sinxcosx [from sin(a + b) = sin a cos b + cos a sin b]:
2sin x cos x - sin x cos x = cos x
sin x cos x - cos x = 0
Factor out cos x:
cos x(sin x - 1) = 0
cos x = 0
x = π/2, 3π/2
sin x = 1
x = π/2
∴ x = π/2, 3π/2