Question

Bài 1: a) Cho cotx=3. Tìm:

\(B=\dfrac{2sin^2x+3sinx.cosx}{1-2cos^2x}\)

b) Cho tanx=-3; \(\dfrac{3\pi}{2}< x< 2\pi\)

Tìm: A=\(\sqrt{10}cosx-2sinx+3\)

 

Answer 1

b: 

3/2pi<x<2pi

=>cosx>0; sin x<0

\(1+tan^2x=\dfrac{1}{cos^2x}\)

=>\(\dfrac{1}{cos^2x}=1+\left(-3\right)^2=10\)

=>cosx=1/căn 10

=>sin x=-3/căn 10

\(A=\sqrt{10}\cdot\dfrac{1}{\sqrt{10}}-2\cdot\dfrac{-3}{\sqrt{10}}+3=4+\dfrac{6}{\sqrt{10}}=\dfrac{4\sqrt{10}+6}{\sqrt{10}}\)

a: cot x=3 nên cosx/sinx=3

=>cosx=3*sinx

\(B=\dfrac{2sin^2x+3sinx\cdot3\cdot sinx}{1-2\cdot\left(3\cdot sinx\right)^2}=\dfrac{11sin^2x}{sin^2x+cos^2x-18sin^2x}\)

\(=\dfrac{11sin^2x}{-17sin^2x+9sin^2x}=\dfrac{-11}{8}\)