a) 2 cos?x - 3 cosx + 1 = 0; 6) 2 sin°x + sinx -1=0

Question

Задача Задача №2209431

А

Аноним

12 ноября 2024 17:55 · публичный вопрос

А

Аноним 12 ноября 2024 17:55

Ответ нейросети

12 ноября 2024 17:55

Let's solve each of the equations step-by-step.

This is a quadratic equation in terms of \( \cos x \). Let's set \( y = \cos x \), then the equation becomes: \[ 2y^2 - 3y + 1 = 0 \]
Now, we will factor this quadratic equation. We look for two numbers that multiply to \( 2 \times 1 = 2 \) and add up to \( -3 \). The factors are \( -1 \) and \( -2 \): \[ 2y^2 - 2y - y + 1 = 0 \] \[ (2y^2 - 2y) + (-y + 1) = 0 \] \[ 2y(y - 1) - 1(y - 1) = 0 \] \[ (2y - 1)(y - 1) = 0 \]
Setting each factor to zero gives: \[ 2y - 1 = 0 \quad \text{or} \quad y - 1 = 0 \] \[ y = \frac{1}{2} \quad \text{or} \quad y = 1 \]
Substituting back \( y = \cos x \):
- For \( \cos x = \frac{1}{2} \): \[ x = \frac{\pi}{3} + 2k\pi \quad \text{or} \quad x = \frac{5\pi}{3} + 2k\pi, \quad k \in \mathbb{Z} \]
- For \( \cos x = 1 \): \[ x = 0 + 2k\pi, \quad k \in \mathbb{Z} \]

This is also a quadratic equation, but in terms of \( \sin x \). Let \( y = \sin x \): \[ 2y^2 + y - 1 = 0 \]
Now, we'll factor this quadratic equation. We need to find two numbers that multiply to \( 2 \times (-1) = -2 \) and add up to \( 1 \). The factors are \( 2 \) and \( -1 \): \[ 2y^2 + 2y - y - 1 = 0 \] \[ (2y^2 + 2y) + (-y - 1) = 0 \] \[ 2y(y + 1) - 1(y + 1) = 0 \] \[ (2y - 1)(y + 1) = 0 \]
Setting each factor to zero gives: \[ 2y - 1 = 0 \quad \text{or} \quad y + 1 = 0 \] \[ y = \frac{1}{2} \quad \text{or} \quad y = -1 \]
Substituting back \( y = \sin x \):
- For \( \sin x = \frac{1}{2} \): \[ x = \frac{\pi}{6} + 2k\pi \quad \text{or} \quad x = \frac{5\pi}{6} + 2k\pi, \quad k \in \mathbb{Z} \]
- For \( \sin x = -1 \): \[ x = \frac{3\pi}{2} + 2k\pi, \quad k \in \mathbb{Z} \]

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