Rozwiąż nierówność sin²x - ½sin2x <0
sin2x=2sinxcosx
sin²x- sinxcosx<0
sinx(sinx-cosx)<0
1.sinx<0 ∧ sinx-cox>0
sinx>cosx
x∈(π+k2π,2π+k2π) x∈(π/4+2kπ,5/4π+2kπ)
x∈(π+k2π,5/4π+k2π)
lub
2
sinx>0 ∧ sinx-cosx<0
sinx<cosx
x∈(0+k2π,π+k2π) x∈(5/4π+k2π,9/4π+k2π)
x∈(2π+k2π,9/4π+k2π)
odp
x∈(π+kπ,5/4π+kπ)
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
sin2x=2sinxcosx
sin²x- sinxcosx<0
sinx(sinx-cosx)<0
1.sinx<0 ∧ sinx-cox>0
sinx>cosx
x∈(π+k2π,2π+k2π) x∈(π/4+2kπ,5/4π+2kπ)
x∈(π+k2π,5/4π+k2π)
lub
2
sinx>0 ∧ sinx-cosx<0
sinx<cosx
x∈(0+k2π,π+k2π) x∈(5/4π+k2π,9/4π+k2π)
x∈(2π+k2π,9/4π+k2π)
odp
x∈(π+kπ,5/4π+kπ)