Помогите решить пожалуйста!!!

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Natasenika13

30 марта 2014 г., 5:19:35 (10 лет назад)

1а)x=+-π/4+2πn
б)1-2sin²x-sinx=0
sinx=a
2a²+a-1=0
D=1+8=9
a1=(-1-3)/4=-1⇒sinx=-1⇒x=-π/2+2πn
a2=(-1+3)/4=1/2⇒sinx=1/2⇒x=(-1)^n*π/6+πn
в)sin²x+sinxcosx-2cos²x=0  /cos²x≠0
tg²x+tgx-2=0
tgx=a
a²+a-2=0⇒a1+a2=-1 U a1*a2=-2⇒
a1=-2⇒tgx=-2⇒x=-arctg2+πn
a2=1⇒tgx=1⇒x=π/4+πn
г)6sin3x/2cos3x/2+5cos²3x/2-5sin²3x/2-4sin²3x/2-4cos²3x/2=0
9sin²3x/2-6sin3x/2cos3x/2-cos²3x/2=0  /cos²3x/2≠0
9tg²3x/2-6tg3x/2-1=0
tg3x/2=a
9a²-6a-1=0
D=36+36=72    √D=6√2
a1=(6-6√2)/18=1/3-√2/3⇒tg3x/2=1/3-√2/3⇒x=2/3arctg(1/3-√2/3)+πn
a2=(6+6√2)/18=1/3+√2/3⇒tg3x/2=1/3+√2/3⇒x=2/3arctg(1/3+√2/3)+πn
2a)-π/2+πn<x≤π/3+πn⇒x∈(-π/2+πn;π/3+πn]
б)3π/4+2πn≤x/2+1/4≤5π/4+2πn
3π/4-1/4+2πn≤x/2≤5π/4-1/4+2πn
3π/2-1/2+4πn≤x≤5π/2-1/2+4πn⇒x∈[3π/2-1/2+4πn;5π/2-1/2+4πn]
3)x=π/2+y
cos(π/2+y)-cosy=-√2
-siny-cosy=-√2
√2/2siny+√2/2cosy=1
sin(y+π/4)=1⇒y+π/4=π/2⇒y=π/2-π/4=π/4⇒x=π/2+π/4=3π/4        (3π/4;π/4)
4)sin²x+sinxcosx-1=0,  sinx≠0
sin²x+sinxcosx-sin²x-cos²x=0
sinxcosx-cos²x=0  /cos²x≠0
tgx-1=0⇒tgx=1⇒x=π/4+πn