y = px+(1/sin(p)) (уравнение)

Найду корень уравнения: y = px+(1/sin(p))

Вы ввели

            1   
y = p*x + ------
          sin(p)

$$y = p x + \frac{1}{\sin{\left(p \right)}}$$

Упростить

Подробное решение

Дано уравнение:
$$y = p x + \frac{1}{\sin{\left(p \right)}}$$
преобразуем:
$$y = p x + \frac{1}{\sin{\left(p \right)}}$$
Раскрываем скобочки в правой части ур-ния

y = 1/sinp + p*x

Получим ответ: y = 1/sin(p) + p*x

Быстрый ответ

       /                 cos(re(p))*sinh(im(p))                        \                  cosh(im(p))*sin(re(p))                        
y1 = I*|- --------------------------------------------------- + im(p*x)| + --------------------------------------------------- + re(p*x)
       |     2            2              2           2                 |      2            2              2           2                 
       \  cos (re(p))*sinh (im(p)) + cosh (im(p))*sin (re(p))          /   cos (re(p))*sinh (im(p)) + cosh (im(p))*sin (re(p))          

$$y_{1} = i \left(\operatorname{im}{\left(p x\right)} - \frac{\cos{\left(\operatorname{re}{\left(p\right)} \right)} \sinh{\left(\operatorname{im}{\left(p\right)} \right)}}{\sin^{2}{\left(\operatorname{re}{\left(p\right)} \right)} \cosh^{2}{\left(\operatorname{im}{\left(p\right)} \right)} + \cos^{2}{\left(\operatorname{re}{\left(p\right)} \right)} \sinh^{2}{\left(\operatorname{im}{\left(p\right)} \right)}}\right) + \operatorname{re}{\left(p x\right)} + \frac{\sin{\left(\operatorname{re}{\left(p\right)} \right)} \cosh{\left(\operatorname{im}{\left(p\right)} \right)}}{\sin^{2}{\left(\operatorname{re}{\left(p\right)} \right)} \cosh^{2}{\left(\operatorname{im}{\left(p\right)} \right)} + \cos^{2}{\left(\operatorname{re}{\left(p\right)} \right)} \sinh^{2}{\left(\operatorname{im}{\left(p\right)} \right)}}$$