Решить уравнения 2sin^2x + sinx - 1 = 0 2cosx^2x - sinx +1 = 0 4sin^2x - cosx - 1 = 0 tg^2x = 2

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Решить уравнения 2sin^2x + sinx - 1 = 0 2cosx^2x - sinx +1 = 0 4sin^2x - cosx - 1 = 0 tg^2x = 2

The_YouTuber 02.03.2026 13:16

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Лебедев Дмитрий Сергеевич · Accepted Answer

Ниже представлено подробное решение четырех тригонометрических уравнений. 1. Решение уравнения $2 \sin^{2} x + \sin x - 1 = 0 2 sine squared x plus sine x minus 1 equals 0$ Это квадратное уравнение относительно $\sin x sine x$ . Введем замену: $t = \sin x t equals sine x$ , где $| t | \leq 1 the absolute value of t end-absolute-value is less than or equal to 1$ . $2 t^{2} + t - 1 = 0 2 t squared plus t minus 1 equals 0$ Находим дискриминант: $D = 1^{2} - 4 \cdot 2 \cdot (-1) = 1 + 8 = 9 = 3^{2} cap D equals 1 squared minus 4 center dot 2 center dot open paren negative 1 close paren equals 1 plus 8 equals 9 equals 3 squared$ Корни уравнения: $t_{1} = \frac{-1 + 3}{4} = \frac{1}{2} t sub 1 equals the fraction with numerator negative 1 plus 3 and denominator 4 end-fraction equals one-half$ $t_{2} = \frac{-1 - 3}{4} = -1 t sub 2 equals the fraction with numerator negative 1 minus 3 and denominator 4 end-fraction equals negative 1$ Обратная замена:

$\sin x = \frac{1}{2} ⟹ x = (-1)^{n} \frac{π}{6} + π n, n \in Z sine x equals one-half ⟹ x equals open paren negative 1 close paren to the n-th power the fraction with numerator pi and denominator 6 end-fraction plus pi n comma n is an element of the integers$ $\sin x = -1 ⟹ x = - \frac{π}{2} + 2 π k, k \in Z sine x equals negative 1 ⟹ x equals negative the fraction with numerator pi and denominator 2 end-fraction plus 2 pi k comma k is an element of the integers$

Ответ: $x = (-1)^{n} \frac{π}{6} + π n x equals open paren negative 1 close paren to the n-th power the fraction with numerator pi and denominator 6 end-fraction plus pi n$ ; $x = - \frac{π}{2} + 2 π k x equals negative the fraction with numerator pi and denominator 2 end-fraction plus 2 pi k$ 2. Решение уравнения $2 \cos^{2} x - \sin x + 1 = 0 2 cosine squared x minus sine x plus 1 equals 0$ (Примечание: в исходном тексте было указано 2cosx^2x, что интерпретируется как $2 \cos^{2} x 2 cosine squared x$ ). Используем основное тригонометрическое тождество $\cos^{2} x = 1 - \sin^{2} x cosine squared x equals 1 minus sine squared x$ : $2 (1 - \sin^{2} x) - \sin x + 1 = 0 2 open paren 1 minus sine squared x close paren minus sine x plus 1 equals 0$ $2 - 2 \sin^{2} x - \sin x + 1 = 0 2 minus 2 sine squared x minus sine x plus 1 equals 0$ $-2 \sin^{2} x - \sin x + 3 = 0 negative 2 sine squared x minus sine x plus 3 equals 0$ $2 \sin^{2} x + \sin x - 3 = 0 2 sine squared x plus sine x minus 3 equals 0$ Пусть $t = \sin x t equals sine x$ : $2 t^{2} + t - 3 = 0 2 t squared plus t minus 3 equals 0$ $D = 1 - 4 \cdot 2 \cdot (-3) = 25 cap D equals 1 minus 4 center dot 2 center dot open paren negative 3 close paren equals 25$ $t_{1} = \frac{-1 + 5}{4} = 1 t sub 1 equals the fraction with numerator negative 1 plus 5 and denominator 4 end-fraction equals 1$ $t_{2} = \frac{-1 - 5}{4} = -1.5 t sub 2 equals the fraction with numerator negative 1 minus 5 and denominator 4 end-fraction equals negative 1.5$ (не подходит, так как $| \sin x | \leq 1 the absolute value of sine x end-absolute-value is less than or equal to 1$ ) Обратная замена: $\sin x = 1 ⟹ x = \frac{π}{2} + 2 π n, n \in Z sine x equals 1 ⟹ x equals the fraction with numerator pi and denominator 2 end-fraction plus 2 pi n comma n is an element of the integers$ Ответ: $x = \frac{π}{2} + 2 π n x equals the fraction with numerator pi and denominator 2 end-fraction plus 2 pi n$ 3. Решение уравнения $4 \sin^{2} x - \cos x - 1 = 0 4 sine squared x minus cosine x minus 1 equals 0$ Заменим $\sin^{2} x sine squared x$ на $1 - \cos^{2} x 1 minus cosine squared x$ : $4 (1 - \cos^{2} x) - \cos x - 1 = 0 4 open paren 1 minus cosine squared x close paren minus cosine x minus 1 equals 0$ $4 - 4 \cos^{2} x - \cos x - 1 = 0 4 minus 4 cosine squared x minus cosine x minus 1 equals 0$ $-4 \cos^{2} x - \cos x + 3 = 0 negative 4 cosine squared x minus cosine x plus 3 equals 0$ $4 \cos^{2} x + \cos x - 3 = 0 4 cosine squared x plus cosine x minus 3 equals 0$ Пусть $t = \cos x t equals cosine x$ : $4 t^{2} + t - 3 = 0 4 t squared plus t minus 3 equals 0$ $D = 1 - 4 \cdot 4 \cdot (-3) = 49 cap D equals 1 minus 4 center dot 4 center dot open paren negative 3 close paren equals 49$ $t_{1} = \frac{-1 + 7}{8} = \frac{3}{4} = 0.75 t sub 1 equals the fraction with numerator negative 1 plus 7 and denominator 8 end-fraction equals three-fourths equals 0.75$ $t_{2} = \frac{-1 - 7}{8} = -1 t sub 2 equals the fraction with numerator negative 1 minus 7 and denominator 8 end-fraction equals negative 1$ Обратная замена:

$\cos x = 0.75 ⟹ x = \pm \arccos (0.75) + 2 π n, n \in Z cosine x equals 0.75 ⟹ x equals plus or minus arc cosine 0.75 plus 2 pi n comma n is an element of the integers$ $\cos x = -1 ⟹ x = π + 2 π k, k \in Z cosine x equals negative 1 ⟹ x equals pi plus 2 pi k comma k is an element of the integers$

Ответ: $x = \pm \arccos (0.75) + 2 π n x equals plus or minus arc cosine 0.75 plus 2 pi n$ ; $x = π + 2 π k x equals pi plus 2 pi k$ 4. Решение уравнения $\tan^{2} x = 2 tangent squared x equals 2$ Извлечем квадратный корень из обеих частей уравнения: $\tan x = \pm \sqrt{2} tangent x equals plus or minus the square root of 2 end-root$ Это распадается на два простейших уравнения:

$\tan x = \sqrt{2} ⟹ x = \arctan (\sqrt{2}) + π n, n \in Z tangent x equals the square root of 2 end-root ⟹ x equals arc tangent open paren the square root of 2 end-root close paren plus pi n comma n is an element of the integers$ $\tan x = - \sqrt{2} ⟹ x = - \arctan (\sqrt{2}) + π k, k \in Z tangent x equals negative the square root of 2 end-root ⟹ x equals negative arc tangent open paren the square root of 2 end-root close paren plus pi k comma k is an element of the integers$

Оба решения можно объединить в одну запись. Ответ: $x = \pm \arctan \sqrt{2} + π n, n \in Z x equals plus or minus arc tangent the square root of 2 end-root plus pi n comma n is an element of the integers$ Могу ли я помочь с решением других систем уравнений или задач по тригонометрии?

Решить уравнения 2sin^2x + sinx - 1 = 0 2cosx^2x - sinx +1 = 0 4sin^2x - cosx - 1 = 0 tg^2x = 2

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