huh? HUH?
Do you want to find solutions to the equation
3sinx - 2cosx = 0 ?
since 3^2+2^2 = 13, this is equivalent to
√13 (3/√13 sinx - 2/√13 cosx) = 0
Now, let y be the angle such that cosy = 3/√13. Then siny = 2/√13. Now you have
√13 (sinx cosy - cosx siny) = 0
√13 sin(x-y) = 0
So, x-y=0 or x-y = π
That is, x-y = kπ for any integer k.
Thus, x = kπ + arcsin(2/√13)
3sinx-2cosx=o; find G s
3 answers
OR
3 sin x - 2 cos x = 0
3 sin x = 2 cos x
3 sin x / cos x = 2
3 tan x = 2
tan x = 2 / 3
x = k π + tan⁻¹ ( 2 / 3 )
x = k π + arctan ( 2 / 3 )
k π + arctan ( 2 / 3 ) and x = k π + arcsin ( 2 / √13 )
is the same solution is written in a different way.
3 sin x - 2 cos x = 0
3 sin x = 2 cos x
3 sin x / cos x = 2
3 tan x = 2
tan x = 2 / 3
x = k π + tan⁻¹ ( 2 / 3 )
x = k π + arctan ( 2 / 3 )
k π + arctan ( 2 / 3 ) and x = k π + arcsin ( 2 / √13 )
is the same solution is written in a different way.
much simpler, Bosnian. I missed that one, fer shure!