решите тригонометрическое уравнение: 4sin^2x-sin2x=3. Created by Nishinoyaa. algebra-ru

решите тригонометрическое уравнение: 4sin^2x-sin2x=3...

Dимасuk

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4sin²x - sin2x = 3

Разложим синус удвоенного аргумента и воспользуемся основным тригонометрическим тождеством:

4sin²x - 2sinxcosx = 3sin²x + 3cos²x

4sin²x - 3sin²x - 2sinxcosx - 3cos²x = 0

sin²x - 2sinxcosx - 3cos²x = 0

Делим всё уравнение на cos²x (cosx ≠ 0).

tg²x - 2tgx - 3 = 0

tg²x - 2tgx + 1 - 4 = 0

(tgx - 1)² - 2² = 0

(tgx - 1 - 2)(tgx - 1 + 2) = 0

(tgx - 3)(tg + 1) = 0

tgx = 3 или tgx = -1

x = atctg3 + πn, n ∈ Z; x = -π/4 + πn, n ∈ Z.

Ответ: x = -π/4 + πn; atctg3 + πn, n ∈ Z.

2 votes Thanks 1

Universalka Минус пропустили в первом ответе

Universalka

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$4Sin^{2}x-Sin2x=3\\\\4Sin^{2}x-2SinxCosx -3*1=0\\\\4Sin^{2}x-2SinxCosx-3(Sin^{2}x+Cos^{2}x)=0\\\\4Sin^{2}x-2SinxCosx-3Sin^{2}x-3Cos^{2}x=0\\\\Sin^{2}x-2SinxCosx-3Cos^{2}x=0$

Разделим почленно на Cos²x ≠ 0 , получим

$\frac{Sin^{2}x}{Cos^{2}x}-\frac{2SinxCosx}{Cos^{2}x}-\frac{3Cos^{2}x}{Cos^{2}x}=0\\\\tg^{2}x-2tgx-3=0\\\\tgx=m\\\\m^{2} -2m-3=0\\\\D=(-2)^{2}-4*(-3)=4+12=16=4^{2}\\\\m_{1}=\frac{2+4}{2}=3\\\\m_{2}=\frac{2-4}{2}=-1\\\\tgx_{1}=3\\\\x_{1} =arctg3+ pin \\\\tgx_{2}=-1\\\\x_{2}=-\frac{\pi}{4}+ pin$

Везде добавить n ∈ z

1 votes Thanks 2

Answers & Comments

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