• Предмет: Математика
  • Автор: vlad30121982
  • Вопрос задан 2 года назад

Нужно решить тригонометрические уравнения​

Приложения:

Ответы

Ответ дал: Miroslava227
1

Ответ:

а

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.

б

 \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.

в

 \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}

г

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}

д

 \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

е

 \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}

ж

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

з

 \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}

и

 \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}

к

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

л

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

Вас заинтересует