Объяснение:
1) cos3x-sin3x=0
(√2/2)cos3x-(√2/2)sin3x=0
cos(π/4)cos3x-sin(π/4)sin3x=0
cos(3x+π/4)=0
3x+π/4=π/2+kπ
3x=π/2-π/4+kπ
3x=π/4+kπ
x=π/12+kπ/3, k∈Z
ответ: x=π/12+kπ/3, k∈Z
2) sin(5x)-√3cos(5x)=0
0,5sin(5x)-0,5√3cos(5x)=0
cos(π/3)sin(5x)-sin(π/3)cos(5x)=0
sin(5x-π/3)=0
5x-π/3=kπ
5x=π/3+kπ
x=π/15+kπ/5, k∈Z
ответ: x=π/15+kπ/5, k∈Z
3) 4sin(x/3)-7cos(x/3)=0
(4/√65)sin(x/3)-(7/√65)cos(x/3)=0
cosα=4/√65; α∈(0;π/2)⇒sinα=7/√65, α=arccos(4/√65)
cosαsin(x/3)-sinαcos(x/3)=0
sin(x/3-α)=0
x/3-α=kπ
x/3=α+kπ
x=3α+3kπ=3arccos(4/√65)+3kπ
ответ:x=3arccos(4/√65)+3kπ
4) 3sin²(x/5)-7sin(x/5)cos(x/5)+4cos²(x/5)=0
3sin²(x/5)/cos²(x/5)-7sin(x/5)cos(x/5)/cos²(x/5)+4cos²(x/5)/cos²(x/5)=0
3tg²(x/5)-7tg(x/5)+4=0; tg(x/5)=y
3y²-7y+4=0
D=49-48=1
y₁=(7-1)/6=1⇒tgx=1⇒x/5=π/4+kπ, x=5π/4+5kπ, k∈Z
y₂=(7+1)/6=4/3⇒tgx=4/3⇒x/5=arctg(4/3)+kπ⇒x=5arctg(4/3)+5π, k∈Z
ответ:x={5π/4+5kπ; 5arctg(4/3)+5π}, k∈Z
№2
1) 7sin²(x/3)-4sin(2x/3)+cos²(x/3)=0
7sin²(x/3)-8sin(x/3)cos(x/3)+cos²(x/3)=0
7sin²(x/3)/cos²(x/3)-8sin(x/3)cos(x/3)/cos²(x/3)+cos²(x/3)/cos²(x/3)=0
7tg²(x/3)-8tg(x/3)+1=0; tg(x/3)=y
7y²-8y+1=0
D=64-28=36
y₁=(8+6)/14=1⇒tgx=1⇒x/3=π/4+kπ, x=3π/4+3kπ, k∈Z
y₂=(8-6)/14=1/7⇒tgx=1/7⇒x/3=arctg(1/7)+kπ⇒x=3arctg(1/7)+3π, k∈Z
ответ:x={3π/4+3kπ; 3arctg(1/7)+3π}, k∈Z
2) (2sinx-cosx)/(cosx+3sinx)=1/4
4(2sinx-cosx)=cosx+3sinx
8sinx-4cosx-cosx-3sinx=0
5sinx-5cosx=0
5(sinx-cosx)=0
sinx=cosx
sinx/cosx=cosx)/cosx
tgx=1
x=π/4+kπ, k∈Z
ответ:x=π/4+kπ, k∈Z
