Решите тригонометрическое уравнение: 1) 4sin^2x-sin2x=3 2) sin2x+8sin^2x=5 3) 10cos^2x-2sin2x=3

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Fansik34324 08.10.2020 21:58

$1) \: \: 4 {sin}^{2} x - sin2x = 3 \\ 4 {sin}^{2} x - sin2x - 3 = 0 \\ 4 {sin}^{2} x - 2sinxcosx - 3( {sin}^{2} x + {cos}^{2} x) = 0 \\ 4 {sin}^{2} x - 2sinxcosx - 3 {sin}^{2} x - 3 {cos}^{2} x = 0 \\ {sin}^{2} x - 2sinxcosx - 3 {cos}^{2} x = 0 \\ \frac{ {sin}^{2} x}{ {cos}^{2} x} - \frac{2sinxcosx}{ {cos}^{2} x} - \frac{3 {cos}^{2} x}{ {cos}^{2} x} = 0 \\ {tg}^{2} x - 2tgx - 3 = 0 \\ tgx = t \\ {t}^{2} - 2t - 3 = 0 \\ d = {b}^{2} - 4ac = 4 - 4 \times ( - 3) = 16 \\ t1 = \frac{2 + 4}{2} = 3 \\ t2 = \frac{2 - 4}{2} = - 1 \\ 1)tgx = 3 \\ x = arctg3 + \pi n \\ 2) tgx = - 1 \\ x = - \frac{\pi}{4} + \pi n$
ответ; arctg3 + pi*n; -pi/4 + pi*n, n € Z.

2)
$sin2x + 8 {sin}^{2} x = 5 \\ sin2x + 8 {sin}^{2} x - 5 = 0 \\ 2sinxcosx + 8 {sin}^{2} x - 5( {sin}^{2} x + {cos}^{2} x) = 0 \\2sinxcosx + 8 {sin}^{2} x - 5 {sin}^{2} x - 5 {cos}^{2} x = 0 \\ 3 {sin}^{2} x + 2sinxcosx - 5 {cos}^{2} x = 0 \\ \frac{3 {sin}^{2} x}{ {cos}^{2} x} + \frac{2sinxcosx}{ {cos}^{2}x } - \frac{5 {cos}^{2} x}{ {cos}^{2}x } = 0 \\ 3 {tg}^{2} x + 2tgx - 5 = 0 \\ tgx = t \\ 3 {t}^{2} + 2t - 5 = 0 \\ d = {b}^{2} - 4ac = 4 - 4 \times 3 \times ( - 5) = 64 \\ t1 = \frac{ - 2 + 8}{2 \times 3} = \frac{6}{6} = 1 \\ t2 = \frac{ - 2 - 8}{2 \times 3} = \frac{ - 10}{6} = - \frac{5}{3} \\ 1)tgx = 1 \\ x = \frac{\pi}{4} + \pi n \\ 2)tgx = - \frac{5}{3} \\ x = - arctg \frac{5}{3} + \pi n$
ответ: pi/4 + pi*n; -arctg5/3 + pi*n, n € Z.

3)
$10 {cos}^{2} x - 2sin2x = 3 \\ 10 {cos}^{2} x - 4sin2x - 3 = 0 \\ 10 {cos}^{2} x - 4sinxcosx - 3( {sin}^{2} x + {cos}^{2} x ) = 0 \\ 10 {cos}^{2} x - 4sinxcosx - 3 {sin}^{2} x - 3 {cos}^{2} x = 0 \\ 7 {cos}^{2} x - 4sinxcosx - 3 {sin}^{2} x = 0 \\ \frac{7 {cos}^{2} x}{ {cos}^{2} x} - \frac{4sinxcosx}{ {cos}^{2}x } - \frac{3 {sin}^{2} x}{ {cos}^{2} x} = 0 \\ 7 - 4tgx - 3 {tg}^{2} x = 0 \\ 3 {tg}^{2} x + 4tgx - 7 = 0 \\ tgx = t \\ 3 {t}^{2} + 4t - 7 = 0 \\ d = {b}^{2} - 4ac = 16 - 4 \times 3 \times ( - 7 ) = 100 \\ t1 = \frac{ - 4 + 10}{2 \times 3} = \frac{6}{6} = 1 \\ t2 = \frac{ - 4 - 10}{2 \times 3} = \frac{ - 14}{6} = - \frac{7}{3} \\ 1)tgx = 1 \\ x = \frac{\pi}{4} + \pi n \\ 2)tgx = - \frac{7}{3} \\ x = - arctg \frac{7}{3} + \pi n$
ответ: pi/4 + pi*n; -arctg7/3 + pi*n, n € Z.

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