有a.maths 唔識做

1.

√2cos 2x = sinx + cosx
=√2(sinxsinx+cosxcosx)
=√2cos(θ-x)
√2cos2x =√2(22/7/4-x)
2x=2n22/7+-(22/7/4-x)

2x=2n(22/7)+(22/7/4-x) or 2x=2n22/7-(22/7/4-x)
2x=2n(22/7)+22/4-x x=2n(22/7)-(22/7/4)
3x=2n(22/7)+(22/7/4)
x=2n(22/7)/3+(22/7)/12

2) sin3x + cos2x = 0
sin3x+sin(90*-2x)=0
sin3x+sin2x=0
2sin(3x+2x/2)cos(3x-2x/2)=0
(2sin5x/2)(cosx/2)=0
(2sin5x/2) = 0 or (cosx/2)=0
sin5x=0 x=2(22/7)n+-(22/7/2)
x=(22/7)n/5
5) tanx = cotx
=tan(22/7/2-x)
x =(22/7)n+(22/7/2-2x)

3)+4)明天補返俾你
(22/7)=180*=3.14(打5到你 圓周符號.你自己change返佢呀)

(1)√2cos 2x = sinx + cosx
√2cos 2x =√2sin(x+π/4)
cos2x=sin(x+π/4)
cos2x=sin(π/4-x)
2x=2nπ(π/4-x),where n is an integer.
2x=2nπ+(π/4-x) or 2x=2nπ-(π/4-x)
3x=2nπ+π/4 or x=2nπ-π/4
∴x=(2nπ)/3+π/12 or x=2nπ-π/4
2) sin3x + cos2x = 0
cos2x=-sin3x
cos2x=cos(π/2+3x)
2x=2nπ(π/2+3x),where n is an integer.
2x=2nπ+(π/2+3x) or 2x=2nπ-(π/2+3x)
-x=2nπ+π/2 or 5x=2nπ-π/2
∴x=-(2nπ+π/2 )or x=(2nπ)/5-π/10





4) 16cos4 x = 1
cos4x=1/16
cosx=1/2 or cosx=-1/2
x=2nππ/3 or x x=2nπ2π/3
5) tanx = cotx
tanx =tan(π/2 –x)
x=nπ+(π/2 –x) ,where n is an integer.
2x= nπ+π/2
x=nπ/2+π/4