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Matematika / Trigonometriya

Trigonometrik tenglamalar

Test: 10 ta savol, vaqt bilan

4 ta oson, 4 ta oʻrta, 2 ta qiyin. Tez va toʻgʻri javob — koʻproq ball.

Testni boshlash

Yoki quyida mashq qiling: variantni tanlang — javob va izoh chiqadi. Ball berilmaydi.

Oson · 5 ta

1.

2sin⁡x=−12\sin x = -1 tenglamaning [0∘; 360∘][0^\circ;\, 360^\circ] dagi eng kichik yechimini (gradusda) toping.

2.

tan⁡x=1\tan x = 1 tenglamaning eng kichik musbat yechimini (gradusda) toping.

3.

sin⁡x=22\sin x = \frac{\sqrt{2}}{2} tenglamaning [0; 2π][0;\, 2\pi] kesmadagi eng kichik yechimini (radianda) toping.

4.

2cos⁡x=1\sqrt{2}\cos x = 1 tenglamaning [0∘; 360∘][0^\circ;\, 360^\circ] dagi eng kichik yechimini (gradusda) toping.

5.

tan⁡x=3\tan x = \sqrt{3} tenglamaning eng kichik musbat yechimini (gradusda) toping.

Oʻrta · 5 ta

1.

sin⁡(2x)=−32\sin(2x) = - \frac{\sqrt{3}}{2} tenglamaning [0; 2π)[0;\, 2\pi) oraliqdagi yechimlari nechta?

2.

cos⁡x=−22\cos x = - \frac{\sqrt{2}}{2} tenglamaning [0; 2π)[0;\, 2\pi) oraliqdagi barcha yechimlari yigʻindisini toping.

3.

sin⁡(2x)=−12\sin(2x) = - \frac{1}{2} tenglamaning [0; 2π)[0;\, 2\pi) oraliqdagi yechimlari nechta?

4.

sin⁡x=12\sin x = \frac{1}{2} tenglamaning [0; 2π)[0;\, 2\pi) oraliqdagi yechimlari nechta?

5.

cos⁡(2x)=22\cos(2x) = \frac{\sqrt{2}}{2} tenglamaning [0; 2π)[0;\, 2\pi) oraliqdagi yechimlari nechta?

Qiyin · 5 ta

1.

2cos⁡2x−cos⁡x−1=02\cos^2 x - \cos x - 1 = 0 tenglamaning [0; 2π)[0;\, 2\pi) oraliqdagi barcha yechimlari yigʻindisini toping.

2.

4sin⁡2x−1=04\sin^2 x - 1 = 0 tenglamaning [0; 2π)[0;\, 2\pi) oraliqdagi yechimlari nechta?

3.

cos⁡2x+cos⁡x=0\cos^2 x + \cos x = 0 tenglamaning [0; 2π)[0;\, 2\pi) oraliqdagi barcha yechimlari yigʻindisini toping.

4.

2cos⁡2x−cos⁡x−1=02\cos^2 x - \cos x - 1 = 0 tenglamaning [0; 2π)[0;\, 2\pi) oraliqdagi yechimlari nechta?

5.

2cos⁡2x−cos⁡x=02\cos^2 x - \cos x = 0 tenglamaning [0; 2π)[0;\, 2\pi) oraliqdagi barcha yechimlari yigʻindisini toping.

Yordam