Question

Помогите решить:
1) 2sin^2x-sinx-2=0
2) sin2x-cosx=0
3) cos7x+cosx=0
4) tgx-2ctgx=0

Answer 1

1.

$2sin^2x-sin x-2=0\D=(-1)^2-4cdot2cdot(-2)=17\sin xneq dfrac{1+sqrt{17}}{4}>1\sin x=dfrac{1-sqrt{17}}{4}Rightarrow boxed{x=(-1)^karcsinfrac{1-sqrt{17}}{4}+pi k, kin Z}$

2.

$sin2x-cos x=0\2sin xcos x-cos x=0\cos x(2sin x-1)=0\cos x=0Rightarrow boxed{x_1=dfrac{pi}{2} +pi n, nin Z}\2sin x-1=0Rightarrow sin x=dfrac{1}{2} Rightarrow boxed{x_2=(-1)^kdfrac{pi}{6} +pi k, kin Z}$

3.

$cos7x+cos x=0\\2cosdfrac{7x+x}{2}cosdfrac{7x-x}{2}=0\\2cos4xcos3x=0\cos4x=0Rightarrow 4x=dfrac{pi}{2} +pi n Rightarrow boxed{x_1=dfrac{pi}{8} +frac{pi n}{4}, nin Z}\cos3x=0Rightarrow 3x=dfrac{pi}{2} +pi n Rightarrow boxed{x_2=dfrac{pi}{6} +frac{pi n}{3}, nin Z }$

4.

$mathrm{tg}x-2mathrm{ctg}x=0 (sin xneq 0; cos xneq 0) \mathrm{tg}x-dfrac{2}{mathrm{tg}x}=0\mathrm{tg}^2x-2=0\mathrm{tg}^2x=2\mathrm{tg}x=pmsqrt{2} \boxed{x=pmmathrm{arctg}sqrt{2}+pi n, nin Z}$