Question

Решите уравнение: cos x = tg x

Answer 1

$cosx=tgx\\\\cosx-tgx=0\\\\cosx-\dfrac{sinx}{cosx}=0\\\\\dfrac{cos^2x-sinx}{cosx}=0\ \ ,\ \ cosx\ne 0\ \ \to \ \ x\ne \dfrac{\pi}{2}+\pi n\ ,\ n\in Z\\\\\\\dfrac{1-sin^2x-sinx}{cosx}=0\ \ \to \ \ \ sin^2x+sinx-1=0\ ,\\\\t=sinx\ ,\ \ -1\leq t\leq 1\ \ ,\ \ \ \ t^2+t-1=0\ \ ,\ \ D=1+4=5\ ,\\\\t_{1,2}=\dfrac{-1\pm \sqrt5}{2}\\\\\\a)\ \ sinx=\dfrac{-1-\sqrt5}{2}\approx -1,6$

$b)\ \ sinx=\dfrac{-1+\sqrt5}{2}\approx 0,6\ \ \to \ \ \ x=(-1)^{n}\cdot arcsin\dfrac{-1+\sqrt5}{2}+\pi k\ ,\ k\in Z\\\\Otvet:\ \ x=(-1)^{n}\cdot arcsin\dfrac{-1+\sqrt5}{2}+\pi k\ ,\ k\in Z\ .$

Answer 2

cosx = tgx;

cosx = sin x/cos x;

cosx ≠0;x≠π/2+πn n∈Z;

cos²x = sinx

-1+sin²x+sinх=0;

sinх=(-1±√(1+4)/2=sinх=(-1±√5)/2:

sinх=(-1-√5)/2; ∅; т.к. IsinхI≤1

sinх=(-1+√5)/2;

х=(-1)ⁿarcsin((-1+√5)/2) +πn; n∈Z