3. 3sin 4x = (cos 2x – 1)tg x Помогите пожалуйста. Created by barkalayairina. algebra-ru

3. 3sin 4x = (cos 2x – 1)tg x Помогите пожалуйста...

July 2022 1 5 Report

3. 3sin 4x = (cos 2x – 1)tg x
Помогите пожалуйста

Answers & Comments

dasdasfa 3*2sin2x cos2x=-2sin^2 x *(sinx/cosx)
12sinx cosx(1-2sin^2 x)+2sin^3x /cosx=0; cosx≠0
12sinx cos^2 x-24sin^3 x cos^2x+2sin^3 x=0
2sinx(6cos^2 x -12sin^2 xc0s^2x+sin^2 x)=0
sinx=0 ili 6cos^2 x-12sin^2 x cos^2x+sin^2 x=0
x=πn 5cos^2 x-12sin^2 xcos^2 x+1=0
5cos^2 x -12*(1-cos^2 x)cos^2 x+1=0
12cos^4 x-7cos^2 x+1=0
t=cos^2 x; 12t^2-7t+1=0
D=49-48=1; t1=(7-1)/24=1/4t; t2=1/3
cos^2 x=1/4 ili cos^2 x=1/3
cosx=1/2 ili cosx=-1/2; cosx=1/√3 ili cosx=-1/√3
x=+-π/3+2πn ili x=+-(2π/3)+2πn; x=+-arccos(+-1/√3)+2πn что-то много корней! Проверьте

6 votes Thanks 6

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