tan x=-2
=>sin x/cosx=-2
=>sin x=-2*cosx
\(1+tan^2x=\dfrac{1}{cos^2x}\)
=>\(\dfrac{1}{cos^2x}=1+2=3\)
=>\(cos^2x=\dfrac{1}{3}\)
\(H=\dfrac{sin^3x+5\cdot cos^3x}{3\cdot sinx-2\cdot cosx}\)
\(=\dfrac{\left(-2\cdot cosx\right)^3+5\cdot cos^3x}{3\cdot\left(-2\right)\cdot cosx-2\cdot cosx}\)
\(=\dfrac{-8\cdot cos^3x+5\cdot cos^3x}{-6\cdot cos-2\cdot cosx}=\dfrac{-3\cdot cos^3x}{-8\cdot cosx}=\dfrac{3}{8}\cdot cos^2x\)
=3/8*1/3
=1/8
Đúng 0
Bình luận (0)
