2sin^2x-5sin 2x cos 2x+2cos^2x

Question

Алгебра

Лека

9 апреля, 23:17

2sin^2x-5sin 2x cos 2x+2cos^2x

+1

Ответы (1)

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Лин · Answer 1

2sin^2x-5sin 2x cos 2x+2cos^2x=0; 2tq²2x - 5tq2x + 2 = 0;

tq²2x - (1/2+2) tq2x + 1 = 0;

[ tq2x = 1/2; tq2x = 2.

[ 2x=arctq (1/2) + πn; [ 2x=arctq2 + πn, n∈Z.

x₁ = (1/2) * arctq (1/2) + (π/2) * n, n∈Z; x₂ = (1/2) * arctq2 + (π/2) * n, n∈Z.

ответ : (1/2) * arctq (1/2) + (π/2) * n, (1/2) * arctq2 + (π/2) * n, n∈Z.