Question

9cos2x-4cos^2x=11sin2x+9
помогите решить ,10 класс алгебра ​

Answer 1

Ответ:

$9( \cos^{2} (x) - \sin^{2} (x) ) - 4 \cos ^{2} (x) = 11 \times 2 \sin(x) \cos(x) + 9 \\ 9 \cos^{2} (x) - 9 \sin^{2} (x) - 4 \cos^{2} (x) = 22 \sin(x) \cos(x) + 9 \\ 5 \cos^{2} (x) - 9 \sin ^{2} (x) - 2 2 \sin(x) \cos(x) = 9 \\ 5 \cos ^{2} (x) - 9(1 - \cos^{2}(x) ) - 22 \sin(x) \cos(x) = 9 \\ 5 \cos^{2}(x) - 9 + 9 \cos^{2}(x) - 22 \sin(x) \cos(x) = 9 \\ 14 \cos^{2}(x) - 22 \sin(x) \cos(x) = 18 \\ 14 \cos^{2}(x) - 22 \sin(x) \cos(x) - 18( \sin^{2}(x) + \cos^{2}(x)) = 0 \\ 14 \cos^{2}(x) - 22 \sin(x) \cos(x) - 18 \sin^{2}(x) - 18 \cos^{2}(x) = 0 \\ - 4 \cos^{2}(x) - 22 \sin(x) \cos(x) - 18 \sin^{2}(x) = 0 \\ \frac{ - 4 \cos^{2}(x) }{ \cos^{2} (x) } - \frac{22 \sin(x) \cos(x) }{ \cos^{2}(x) } - \frac{18 \sin^{2}(x) }{ \cos^{2}(x) } = 0 \\ - 4 - 22 \tan(x) - 18 \tan^{2}(x) = 0$

Заменим tan(x)=t

$- 4 - 22t - 18 {t}^{2} = 0 \\ 9 {t}^{2} + 11t + 2 = 0$

D=121-72=49

$t = \frac{ - 11 \pm \sqrt{49} }{2 \times 9} = \frac{ - 11 \pm7}{18}$

$t_{1} = \frac{ - 11 - 7}{18} = - 1 \\ t _{2} = \frac{ - 11 + 7}{18} = - \frac{2}{9}$

$\tan(x) = - 1 \\ x = - \frac{3\pi}{4} + \pi k,k \in \Z$

$\tan(x) = - \frac{2}{9} \\ x = - \arctan( \frac{2}{9} ) + \pi n, n \in \Z$