Математика: Нужно решить тригонометрические уравнения

аn принадлежит Z. бn принадлежит Z. вгдежзикл...

Нужно решить тригонометрические уравнения

Ответы

Юля5451 12.05.2021 14:40

$35 \cos(x) + 7 \sin {}^{2} (x) + 7\cos {}^{2} (x) = 6 \\ 35 \cos(x) + 7( \sin {}^{2} (x ) + \cos {}^{2} (x)) = 6 \\ 35 \cos(x) + 7 = 6 \\ 35 \cos(x) = - 1 \\ \cos(x) = - \frac{1}{35} \\ x = \pi\pm \: arccos( \frac{1}{35} ) + 2\pi \: n$

n принадлежит Z.

$\sqrt{2} \cos( \frac{\pi}{4} + \frac{45x}{2} ) - \cos( \frac{45x}{2} ) = 1 \\ \sqrt{2} ( \cos( \frac{\pi}{4} ) \cos( \frac{45x}{2} ) - \sin( \frac{\pi}{4} ) \sin( \frac{45x}{2} ) ) - \cos( \frac{45x}{2} ) = 1 \\ \sqrt{2} ( \frac{ \sqrt{2} }{2} \cos( \frac{45x}{2} ) - \frac{ \sqrt{2} }{2} \sin( \frac{45x}{2} ) ) - \cos( \frac{45x}{2} ) = 1 \\ \cos( \frac{45x}{2} ) - \sin( \frac{45x}{2} ) - \cos( \frac{45x}{2} ) = 1 \\ \sin( \frac{45x}{2} ) = - 1 \\ \frac{45x}{2} = - \frac{\pi}{2} + 2\pi \: n \\ x = - \frac{\pi}{45} + \frac{4\pi \: n}{45}$

n принадлежит Z.

$\cos {}^{2} ( \frac{2x}{7} ) = \frac{3}{4} \\ \cos( \frac{2x}{7} ) = \pm \frac{ \sqrt{3} }{2} \\ \frac{2x}{7} = \pm \frac{\pi}{3} + \pi \: n \\ x = \pm \frac{7\pi}{6} + \frac{7\pi \: n}{2}$

$5 \sin {}^{2} (x) + 25 \sin(x) = 0 \\ 5 \sin(x) ( \sin(x) + 5) = 0 \\ \\ \sin(x) = 0 \\ x = 2\pi \: n \\ \\ \sin(x) = - 5 \\ \text{корней нет}$

$\sin {}^{2} (x) - 4 \sin(x) - 5 = 0 \\ \\ \sin(x) = t \\ \\ t {}^{2} - 4 t - 5 = 0 \\ d = 16 + 20 = 36 \\ t1 = \frac{4 + 6}{2} = 5 \\ t2 = - 1 \\ \\ \sin(x) = 5 \\ \text{корней нет} \\ \\ \sin(x) = - 1 \\ x2 = - \frac{\pi}{2} + 2 \pi \: n$

$\sin {}^{2} (x) + 7 \cos(x) + 29 = 0 \\ 1 - \cos {}^{2} (x) + 7 \cos(x) + 29 = 0 \\ \cos {}^{2} (x) - 7\cos(x) - 30 = 0 \\ D = 49 + 120 = 169 \\ \\ \cos(x) = \frac{7 + 13}{2} = 1 0\\ \text{корней нет} \\ \\ \cos(x) = - 3 \\ \text{корней нет}$

$\sin(x) + 35 \cos(x) = 0 \\ \sin(x) = - 35 \cos(x) | \\ \cos(x) \ne0 \\ tgx = - 35 \\ x = - arctg(35) + \pi \: n$

$\sin( \frac{5x}{2} ) + \cos( \frac{5x}{2} ) = 1 \: \: \: | \times \frac{ \sqrt{2} }{2} \\ \frac{ \sqrt{2} }{2} \sin( \frac{5x}{2} ) + \frac{ \sqrt{2} }{2} \cos( \frac{5x}{2} ) = \frac{ \sqrt{2} }{2} \\ \cos( \frac{\pi}{4} ) \sin( \frac{5x}{2} ) + \sin( \frac{\pi}{4} ) \cos( \frac{5x}{2} ) = \frac{ \sqrt{2} }{2} \\ \sin( \frac{5x }{2} + \frac{\pi}{4} ) = \frac{ \sqrt{2} }{2} \\ \\ \frac{5x1}{2} + \frac{\pi}{4} = \frac{\pi}{4} + 2\pi \: n \\ \frac{5x1}{2} = 2\pi \: n \\ x1 = \frac{4\pi \: n}{5} \\ \\ \frac{5x2}{2} + \frac{\pi}{4} = \frac{3\pi}{4} + 2 \pi \: n \\ \frac{5x2}{2} = \frac{\pi}{2} + 2\pi \: n \\ x2 = \frac{\pi}{5} + \frac{4\pi \: n}{5}$

$\sin {}^{2} ( \frac{2x}{5} ) - 2 \sin( \frac{2x}{5} ) \cos( \frac{2x}{5} ) - 3 \cos {}^{2} ( \frac{2x}{5} ) = 0 \\ | \div \cos {}^{2} ( \frac{2x}{5} ) \ne0 \\ {tg}^{2} ( \frac{2x}{5} ) - 2tg( \frac{2x}{5} ) - 3 = 0 \\ d = 4 + 12 = 16 \\ \\ tg( \frac{2x}{5} ) = \frac{2 + 4}{2} = 3 \\ \frac{2x}{5} = arctg(3) + \pi \: n \\ x1 = \frac{5}{2} arctg(3)+ \frac{5\pi \: n}{2} \\ \\ tg( \frac{2x}{5} ) = - 1 \\ \frac{2x}{5} = - \frac{\pi}{4} + \pi \: n \\ x2 = - \frac{5\pi}{8} + \frac{5\pi \: n}{2}$

$2 \sin {}^{2} (x) = 1 - \cos(x) \\ 2 - 2 \cos {}^{2} (x) + \cos(x) - 1 = 0 \\ 2 \cos {}^{2} (x) - \cos( x) - 1 = 0 \\ D= 1 + 8 = 9 \\ \\ \cos(x) = \frac{1 + 3}{4} = 1 \\ x1 = 2\pi \: n \\ \\ \cos(x) = - \frac{1}{2} \\ x2 = \pm \frac{2\pi}{3} + 2\pi \: n$

$3 \sin {}^{2} (x) - \sin(2x) - \cos {}^{2} (x) = 2 \\ 3 \sin {}^{2} (x) - 2 \sin(x) \cos(x) - \cos {}^{2} (x) = 2 \sin {}^{2} (x) + 2 \cos {}^{2} (x) \\ \sin {}^{2} (x) - 2 \sin(x) \cos(x) - 3 \cos {}^{2} (x) = 0 \\ | \div \cos {}^{2} (x) \ne0 \\ \\ {tgx}^{2} - 2 tgx - 3 = 0 \\ D= 4 + 12 = 16 \\ \\ tgx = \frac{2 + 4}{2} = 3 \\ x1 = arctg(3) + \pi \: n\\ \\ tgx = - 1 \\ x2 = - \frac{\pi}{4} + \pi \: n$