Question

4 tg x – 6ctg x + 5 = 0

Answer 1

Дано уравнение:

$4 \tg(x) - 6 \ctg(x) + 5 = 0$

ОДЗ:

$x≠ \frac{\pi}{2} + \pi l, \: l \in Z$

Упрощаем уравнение:

$4 \tg(x) - \frac{6}{ \tg(x) } + 5 = 0 \\ 4 \tg^{2} (x) + 5 \tg(x) - 6 = 0$

Заменим tgx = t. Тогда наше уравнение будет иметь более упрощённый вид:

$4 {t}^{2} + 5t - 6 = 0 \\ D = 25 + 16 \times 6 = 121 = {11}^{2} \\ t_{1} = \frac{ - 5 + 11}{8} = \frac{3}{4} \\ t_{2} = \frac{ - 5 - 11}{8} = - 2$

Проводим обратную замену:

$1) \: \tg(x) = \frac{3}{4} \\ x = \arctg(0.75) + \pi n , \: n \in Z \\ 2) \: \tg(x) = - 2 \\ x = - \arctg(2) + \pi m, \: m \in Z$

Эти два корня и являются решением.

Answer 2

To solve the trigonometric equation 4tg(x) - 6ctg(x) + 5 = 0, we'll first need to rewrite the equation in terms of a single trigonometric function. Since cotangent (ctg) is the reciprocal of tangent (tg), we can use the identity:

ctg(x) = 1/tg(x)

Now, our equation becomes:

4tg(x) - 6(1/tg(x)) + 5 = 0

To make the equation simpler, let's substitute u = tg(x):

4u - 6(1/u) + 5 = 0

Now, we'll solve for u. To do that, let's find a common denominator and get the equation in a quadratic form:

Multiply the entire equation by u:

4u^2 - 6 + 5u = 0

Rearrange the terms to form a quadratic equation:

4u^2 + 5u - 6 = 0

Now, we can factor this quadratic equation:

(4u + 3)(u - 2) = 0

Now, set each factor to zero and solve for u:

1. 4u + 3 = 0 4u = -3 u = -3/4

2. u - 2 = 0 u = 2

Now that we have the values of u, we can find the values of x using the original substitution u = tg(x):

1. u = -3/4 tg(x) = -3/4

2. u = 2 tg(x) = 2

To find x, take the inverse tangent (tan^(-1)) of each value:

1. x = tan^(-1)(-3/4)

2. x = tan^(-1)(2)

So, the solutions for the equation 4tg(x) - 6ctg(x) + 5 = 0 are:

1. x = tan^(-1)(-3/4)

2. x = tan^(-1)(2)

Keep in mind that there might be other solutions, as trigonometric equations often have multiple periodic solutions.