don't like fractions, so how about Ø = x/2 or x = 2Ø
then your equation becomes
cosØ - sin 2Ø = 0
cosØ - 2sinØcosØ = 0
cosØ(1 - 2sinØ) = 0
cosØ = 0 or sinØ = 1/2
Ø = π/2, 3π/2 ---> x = π, 3π
or
Ø = π/6, 5π/6, ---> x = π/3, 5π/3
in your given interval:
x = π, π/3, 5π/3
or
recall cos 2A = 2cos^2 A - 1 or 1 - 2sin^2 x
so if
cos (x/2) = sinx
squaring both sides brings that equation into play. A messier way of doing it.
Also remember since squaring has taken place, all answers must be checked
Find all solutions of the given equation in the interval
[0, 2pi)
cos x/2 - sin x = 0
Hi, I am struggling with this question. Can anybody help me please? Thanks!
1 answer