Question

математика помогите пожалуйста
80 балов даю​

Answer 1

Объяснение:

2.$\frac{2sin^2\alpha -1}{sin\alpha +cos\alpha } +cos\alpha =\frac{sin^2\alpha +sin^2\alpha -1+(sin\alpha +cos\alpha)*cos\alpha }{sin\alpha +cos\alpha} =\frac{sin^2\alpha -(1-sin^2\alpha)+sin\alpha*cos\alpha +cos^2\alpha }{sin\alpha +cos\alpha} =\\=\frac{sin^2\alpha -cos^2\alpha +sin\alpha *cos\alpha +cos^2\alpha }{{sin\alpha +cos\alpha} } =\frac{sin^2\alpha +sin\alpha *cos\alpha }{{sin\alpha +cos\alpha} }=\frac{sin\alpha *(sin\alpha +cos\alpha )}{sin\alpha +cos\alpha } =sin\alpha .$3.

$sin^2x-4*sinx*cosx+3*cos^2x=0\ |:cosx\ \ (cosx\neq 0\ \ \ \ x\neq \pi n)\\tg^2x-4tgx+3=0\\tgx=v\ \ \ \ \Rightarrow\\v^2-4v+3=0\\D=4 \ \ \ \ \sqrt{D}=2\\t_1=tgx= 1\ \ \ \ x_1=\frac{\pi }{4}+\pi n,\ n\in\mathbb Z \\t_2=tgx=3 \ \ \ \ x_2=arctg(3)+\pi n,\ n\in\mathbb Z.$

Ответ: x₁=π/4+πn,  x₂=arctg(3)+πn.

4.

$sin^4\alpha -cos^4\alpha =(sin^2\alpha -cos^2\alpha )*(sin^2\alpha +cos^2\alpha )=(sin^2\alpha -cos^2\alpha )*1=\\=sin^2\alpha -cos^2\alpha .$