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mrhunter913
@mrhunter913
August 2022
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решить уравнения:
№120 под б, г
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niikaLs
Напишите это уравнение здесь.
3 votes
Thanks 1
mrhunter913
б) 4cos^2x-3sinx=3, г) 3tg^2x+ctg^2x=4
niikaLs
Б) 4cos^2x - 3sinx - 3 = 0
4(1 - sin^2x) - 3sinx - 3 = 0
4 - 4sin^2x - 3sinx - 3 = 0
- 4sin^2x - 3sinx + 1 = 0 // : (-1)
4sin^2x + 3sinx - 1 = 0
Г)3tg²x+1/tg²x=4
tg²x=y;3y+1/y-4=0
3y²-4y+1=0;y₁=1;y₂=⅓
1) tg²x=1;tgx=±1;x=±π/4+πn,n∈Z
2) tg²x=⅓;tgx=±1/√3;x=±π/6+πk,k∈Z.
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Answers & Comments
4(1 - sin^2x) - 3sinx - 3 = 0
4 - 4sin^2x - 3sinx - 3 = 0
- 4sin^2x - 3sinx + 1 = 0 // : (-1)
4sin^2x + 3sinx - 1 = 0
Г)3tg²x+1/tg²x=4
tg²x=y;3y+1/y-4=0
3y²-4y+1=0;y₁=1;y₂=⅓
1) tg²x=1;tgx=±1;x=±π/4+πn,n∈Z
2) tg²x=⅓;tgx=±1/√3;x=±π/6+πk,k∈Z.