1. 3 cos 2x - 5 cos x = 1 2. cos 3x - cos 5x= sin 4x 3. 2 sin x - cos x =0 4. 3 sin^2x + sin x cos x- 4cos ^2x

Question

Алгебра: 1. 3 cos 2x - 5 cos x = 1 2. cos 3x - cos 5x= sin 4x 3. 2 sin x - cos x =0 4. 3 sin^2x + sin x cos x- 4cos ^2x =0

Vzinko42 · Answer

1.3cos2x-5cosx=13(cos²x-sin²x)-5cosx-1=03cos²x-3sin²x-5cosx-1=03cos²x-3(1-cos²x)-5cosx-1=03cos²x-3+3cos²x-5cosx-1=06cos²x-5cosx-4=0cosx=t ; |t|≤16t²-5t-4=0D=25-4*6*(-4)=25+96=121t1=(5+11)/12=16/12(не удовл. усл |t|≤1)t2=(5-11)/12=-6/12=-1/2cosx=-1/2x=+-2π/3+2πn; n∈zответ:+-2π/3+2πn; n∈z﻿...

1. 3 cos 2x - 5 cos x = 1 2. cos 3x - cos 5x= sin 4x 3. 2 sin x - cos x =0 4. 3 sin^2x + sin x cos x- 4cos ^2x =0

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