Question

Помогите с вариантом 1

1) sin\frac{x}{3} =-\frac{1}{2} [0; 3\pi ]

2) 3cosx -cos^{2}x =0

3) 6sin^{2}x-sinx=1

4) 4sinx+5cosx=4

5) sin^{4}x+cos^{4}x=cos^{2}2x+\frac{1}{4}[/tex]

Answer 1

1)3cos²x-5cosx-8=0

cosx=a

3a²-5a-8=0

D=25+96=121

a1=(5-11)/6=-1⇒cosx=-1⇒x=π+2πn,n∈z

a2=(5+11)/6=2 2/3>1 нет решения

2)8cos^2x-14sinx+1=0

8-8sin²x-14sinx+1=0

sinx=a

8a²+14a-9=0

D=196+288=484

a1=(-14-22)/16=-2,25<-1 нет решения

a2=(-14+22)/16=1/2⇒sinx=1/2⇒x=(-1)^n*π/6+πn,n∈z

3)5sin^2x+14 sinxcosx+8cos^2x=0/cos²x

5tg²x+14tgx+8=0

tgx=a

5a²+14a+8=0

D=196-160=36

a1=(-14-6)/10=-2⇒tgx=-2⇒x=-arctg2+πn,n∈z

a2=(-14+6)/10=-0,8⇒tgx=-0,8⇒x=-arctg0,8+πk,k∈z

4)2tgx-9ctgx +3=0

2tgx-9/tgx+3=0

2tg²x+3tgx-9=0,tgx≠0

tgx=a

2a²+3a-9=0

D=9+72=81

a1=(-3-9)/4=-3⇒tgx=-3⇒x=-arctg3+πn,n∈z

a2=(-3+9)/4=1,5⇒tgx=1,5⇒x=arctg1,5+πk,k∈z

5)sin^2x-5cos^2x=2sin2x

sin²x-5cos²x-4sinxcosx=0/cos²x

tg²x-4tgx-5=0

tgx=a

a²-4a-5=0

a1+a2=4 U a1*a2=-5

a1=-1⇒tgx=-1⇒x=-π/4+πn,n∈z

a2=5⇒tgx=5⇒x=arctg5+πk,k∈z

6)5cos2x+5=8sin2x-6sin^2x

5cos²x-5sin²x+5sin²x+5cos²x-16sinxcosx+6sin²x=0/cos²x

6tg²x-16tgx+10=0

tgx=a

3a²-8a+5=0

D=64-60=4

a1=(8-2)/6=1⇒tgx=1⇒x=π/4+πn,n∈z

a2=(8+2)/6=5/3⇒tgx=5/3⇒x=arctg5/3+πk,k∈z