Question

4tg^2x+11/sin(3п/2-x)+10=0

Answer 1

$4tg^2x+ \frac{11}{\sin( \frac{3\pi}{2}-x) } +10=0 \\ 4tg ^2x- \frac{11}{\cos x} +10=0|\cdot \cos^2x \\ 4\sin^2x-11\cos x+10\cos^2x=0 \\ 4(1-\cos^2x)-11\cos x+10\cos^2x=0 \\ 4-4\cos^2x-11\cos x+10\cos^2x=0 \\ 6\cos^2x-11\cos x+4=0$

Пусть $\cos x=t\,\,\, (|t| \leq 1)$
$6t^2-11t+4=0 \\ D=b^2-4ac=121-96=25 \\ t_1= \frac{11+5}{12} = \frac{16}{12} \\ t_2= \frac{11-5}{12} = \frac{1}{2}$
Корень t=(11+5)/12 не удовлетворяет условие при $|t| \leq 1$

Обратная замена

$\cos x= \frac{1}{2} \\ x=\pm \frac{\pi}{3} +2\pi n,n \in Z$

Answer 2

To solve the equation:

$\frac{4tg^2x + 11}{\sin\left(\frac{3\pi}{2} - x\right)} + 10 = 0$

Let's simplify it step by step:

First, we can rewrite $tg^2x$ as $\tan^2x$ (assuming you meant the tangent function).

$\frac{4\tan^2x + 11}{\sin\left(\frac{3\pi}{2} - x\right)} + 10 = 0$

Next, we'll work on simplifying the expression in the denominator. Using the identity $\sin(\frac{\pi}{2} - \theta) = \cos(\theta)$:

$\frac{4\tan^2x + 11}{\cos(x)} + 10 = 0$

To eliminate the fraction, we'll multiply both sides of the equation by $\cos(x)$:

$4\tan^2x + 11 + 10\cos(x) = 0$

Now, the equation is in a form that we can work with. However, finding an exact solution for this equation might be challenging. Depending on the specific requirements of your problem, you might need to use numerical methods or approximations to find an approximate solution.

If you're looking for a numerical solution, you can use numerical methods such as Newton's method or the bisection method to find approximate values of $x$ that satisfy the equation.