Question

Срочно нужно решение

Answer 1

Ответ:

1.

$2 { \cos }^{2} (x) + \cos(x) - 1 = 0$

замена:

$\cos(x) = t \\ \\ 2 {t}^{2} + t - 1 = 0 \\ d = 1 + 8 = 9 \\ t1 = \frac{ - 1 + 3}{4} = \frac{1}{2} \\ t2 = - 1 \\ \\ \cos(x) = \frac{1}{2} \\ x1 = + - \frac{\pi}{3} + 2\pi \: n \\ \\ \cos(x) = - 1 \\ x2 = \pi + 2\pi \: n$

n принадлежит Z.

2.

$2 { \sin }^{2} (x) + \sin( x) - 6 = 0 \\ \\ \sin(x) = t \\ \\ 2 {t}^{2} + t - 6 = 0 \\ d = 1 + 48 = 49 \\ t1 = \frac{ - 1 + 7}{4} = \frac{3}{2} \\ t2 = - 2 \\ \\ \sin(x) = 1.5 \\ \sin(x) = - 2$

в обоих случаях нет корней, так как

$- 1 \leqslant \sin(x) \leqslant 1$

3.

${tg}^{2} x - 3tgx + 2 = 0 \\ \\ tgx = t \\ \\ {t}^{2} - 3t + 2 = 0 \\ d = 9 - 8 = 1 \\ t1 = \frac{3 + 1}{2} = 2 \\ t2 = 1 \\ \\ tgx = 2 \\ x1 = arctg(2) + \pi \: n \\ \\ tgx = 1 \\ x2 = \frac{\pi}{4} + \pi \: n$

n принадлежит Z.

4.

$2 { \cos}^{2} (x) - \sin(x) + 1 = 0 \\ 2 - 2 { \sin }^{2} ( x) - \sin(x) + 1 = 0 \\ 2 { \sin}^{2} (x) + \sin(x) - 3 = 0 \\ \\ \sin(x) = t \\ 2 {t}^{2} + t - 3 = 0 \\ d = 1 + 24 = 25 \\ t1 = \frac{ - 1 + 5}{4} = 1 \\ t2 = - 1.5 \\ \\ \sin(x) = 1 \\ x = \frac{\pi}{2} + 2\pi \: n \\ \\ x2 = - 1.5$

нет корней

Ответ:

$x = \frac{\pi}{2} + 2\pi \: n$

5.

$4 { \sin }^{2} (x) - \cos(x) - 1 = 0 \\ 4 - 4 { \cos }^{2} (x) - \cos(x) - 1 = 0 \\ 4 { \cos }^{2} (x) + \cos(x) - 3 = 0 \\ \\ \cos(x) = t \\ \\ 4 {t}^{2} + t - 3 = 0 \\ d = 1 + 48 = 49 \\ t1 = \frac{ - 1 + 7}{8} = \frac{3}{4} = 0.75 \\ t2 = - 1 \\ \\ \cos(x) = 0.75 \\ x1 = + - arccos(0.75) + 2\pi \: n \\ \\ \cos(x) = - 1 \\ x2 = \pi + 2\pi \: n$

n принадлежит Z.