Forum serwisu

Odp. C

Niech \(x_n=b_1 \cdot q^{n-1} \\ x_{n+1}=b_1 \cdot q^n\)

\(a_{n+1} - a_n =log x_{n+1}-log x_n=log(b_1 \cdot q^n) - log(b_1 \cdot q^{n-1})=log \frac{ logb_1 \cdot q^n }{ logb_1 \cdot q^{n-1} } =logq=const\)
Zatem ciąg a_n jest arytmetyczny.

tg^3 x -tg^2x-3tgx+3=0 \\ tg^2x(tgx-1)-3(tgx-1)=0 \\ (tgx-1)(tg^2x-3)=0 \\ (tgx-1)(tgx- \sqrt{3} ) (tgx+ \sqrt{3} ) =0 \\ tgx=1 \vee tgx= \sqrt{3} \vee tgx=- \sqrt{3} \\ (x= \frac{π}{4}+kπ \vee x= \frac{π}{3}+kπ \vee x=- \frac{π}{3}+kπ ) \wedge x \in (- \frac{π}{2} , \frac{3π}{2} ) \So \\ x= \frac{...

d) 4sin \frac{x}{2}+ cosx =3 \\ 4 sin \frac{x}{2}+1-2 sin ^2 \frac{x}{2} =3 \\ -2 sin^2 \frac{x}{2} +4 sin \frac{x}{2}-2=0 \\ sin^2 \frac{x}{2} -2 sin \frac{x}{2}+1 =0 \\ (sin \frac{x}{2}-1)^2=0 \\sin \frac{x}{2}-1=0 \\ sin \frac{x}{2}=1 \\ \frac{x}{2} = \frac{π}{2} +2kπ, \ k \in C \\ x=π+4kπ, k \in...

c) sinx \cdot tg \frac{x}{2}=1 \quad \quad \quad \quad \quad \frac{x}{2} \neq \frac{π}{2} +kπ, \ k \in C \So x \neq π+2kπ, \ k \in C \\ 2 sin \frac{x}{2} cos \frac{x}{2} \cdot \frac{ sin \frac{x}{2} }{ cos \frac{x}{2} }=1 \\ 2 sin^2 \frac{x}{2} =1 \\ sin^2 \frac{x}{2} = \frac{1}{2} \\ sin \frac{x}{2...

b) 1+cosx+cos \frac{x}{2}=0 \\ 1+2cos^2 \frac{x}{2}-1+cos \frac{x}{2}=0 \\ 2cos^2 \frac{x}{2}+cos \frac{x}{2}=0 \\ cos \frac{x}{2}(2cos \frac{x}{2}+1)=0 \\ cos \frac{x}{2}=0 \vee cos \frac{x}{2}= - \frac{1}{2} \\ \frac{x}{2}= \frac{π}{2} +kπ \vee \frac{x}{2}= \frac{2π}{3} +2kπ \vee\frac{x}{2}= - \fr...

a) 2cosx+3=4cos \frac{x}{2} \\ 2 cos2 \frac{x}{2} +3=4cos \frac{x}{2} \\ 2(cos^2 \frac{x}{2} -1)+3=4cos \frac{x}{2} \\ 4cos^2 \frac{x}{2} -4cos \frac{x}{2} +1=0 \\ (2cos \frac{x}{2} -1)^2=0 \\ 2cos \frac{x}{2} -1= 0 \\ cos \frac{x}{2} = \frac{1}{2} \\ \frac{x}{2}= \frac{π}{3} +2kπ \vee \frac{x}{2}= ...

4sin^3 x - 8sin^2 x - sinx +2 =0 4sin^2x(sinx-2)-(sinx-2)=0 \\ (sinx-2)(4sin^2x-1)=0 \\ sinx=2 \vee sin^2x= \frac{1}{4} \\ x \in \emptyset \vee sinx= \frac{1}{2} \vee sinx=- \frac{1}{2} \\ \quad \quad x= \frac{π}{6}+2kπ \vee x= \frac{5π}{6}+2kπ \vee x=- \frac{π}{6}+2kπ \vee x=- \frac{5π}{6}+2kπ, \ ...

2sin^3 x -3sinx cosx = 0 sinx(2sin^2x-3cosx)=0 \\ sinx(2(1-cos^2x)-3cosx)=0 \\ sinx(-2cos^2x-3cosx+2)=0 \\ sinx=0 \vee -2cos^2x-3cosx+2=0 \\ \quad \quad \quad \quad podstawienie: \ cosx=t \wedge t \in <-1;1> \\ \quad \quad \quad \quad -2t^2-3t+2=0 \\ \quad \quad \quad \quad \Delta =25 \\ \quad \qua...

katie12 pisze:\(ctg^4 x - 2ctg^2 x - 3=0\)

\(x \neq kπ, \ k \in C \\ podst. \ ctg^2x= t \wedge t \in \ge 0 \\ t^2-2t-3=0 \\ \Delta =16 \\ t_1=-1 \\ t_2= 3 \\ ctg^2x=3 \\ ctgx= \sqrt{3} \vee ctgx=- \sqrt{3} \\ x= \frac{π}{6}+kπ \vee x= \frac{5π}{6} +kπ, \ k \in C\)

f(x)= \sqrt{2} sin3x + \sqrt{6} cos3x \\ f(x)=2 \sqrt{2} ( \frac{1}{2} sin3x+ \frac{ \sqrt{3} }{2} cos3x) \\ f(x)=2 \sqrt{2} ( cos \frac{π}{3} sin3x+sin \frac{π}{3} cos3x) \\ f(x)=2 \sqrt{2} sin(3x+ \frac{π}{3}) -1 \le sin(3x+ \frac{π}{3}) \le 1 \\ - 2 \sqrt{2} \le 2 \sqrt{2} sin(3x+ \frac{π}{3}) \...

a)
\(5ctg \alpha + 4cos ^2 \alpha =-4sin ^2 \alpha \\5ctg \alpha =-4sin ^2 \alpha - 4cos ^2 \alpha \\ 5ctg \alpha =-4(sin ^2 \alpha +cos ^2 \alpha ) \\ 5ctg \alpha =-4 \cdot 1 \\ ctg \alpha =- \frac{4}{5}\)

. \begin{cases} \frac{sinx}{cosx} =- \frac{4}{3} \\ sin^2x+cos^2x=1 \end{cases} \\ \begin{cases} sinx = - \frac{4}{3} cosx \\ (- \frac{4}{3} cosx )^2+cos^2x =1\end{cases} \\ \frac{16}{9} cos^2x+cos^2x=1 \\ cos^2x= \frac{9}{25} \\ cosx = - \frac{3}{5} \\ sinx=- \frac{4}{3} \cdot (- \frac{3}{5} )= \f...

Zad.1. Środek koła: S=(30;50) r= \sqrt{400+400} =20 \sqrt{2} P=(n,40) należy do koła, gdy: (n-30)^2+(40-50)^2 \le 800 \\ n^2-60n+900+100 \le 800 \\ n^2-60n+200 \le 0 \\ \Delta =2800 \\ \sqrt{ \Delta} = 20 \sqrt{7} \\ n_1=30-10 \sqrt{7} \\ n_2=30+10 \sqrt{7} \\ n \in <30-10 \sqrt{7} ; 30+10 \sqrt{7} ...

Zad.7.
S=(p,q)
P=(2,1)
O=(0,0)
k=-200

\(J_P^k (0,0)=S \iff \vec{PS}=k \cdot \vec{PO} \\ [p-2,q-1]=-200[-2,-1] \\ [p-2,q-1]=[400,200] \\ p-2=400 \wedge q-1=200 \\ p=402 \wedge q=201 \\S=(402,201)\)

Zad.2. \alpha , \beta , \gamma - kąty trójkąta, zatem zachodzi: \gamma = 180° -( \alpha+ \beta ) \\ sin \gamma =sin ( \alpha + \beta ) sin \gamma= \frac{sin \alpha + sin \beta}{cos \alpha + cos \beta } \\ sin( \alpha + \beta )= \frac{2sin \frac{ \alpha + \beta }{2} cos \frac{ \alpha - \beta }{2} }{2...