Question

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

Answer 1

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