Question

-2 ctg^2x -ctgx-1=0 Тригонометрия 10 класс

Answer 1

Ответ:

2 ctg²x -ctgx-1=0

ctgx= t

2t²-t-1=0

D=1+8=9

√D= 3

t1=1+3/4= 1

t2=1-3/4=-1/2

ctgx= 1 , x= π/4 + πn nєz

ctgx= -1/2 , x= π- arctg1/2 + πn nєz

Answer 2

To solve the trigonometric equation $-2\cot^2(x) - \cot(x) - 1 = 0$, we can make use of a substitution to simplify the expression. Let's introduce a new variable, $u = \cot(x)$. Now the equation becomes a quadratic equation in terms of $u$:

$-2u^2 - u - 1 = 0$

To solve this quadratic equation, we can use factoring or the quadratic formula. In this case, let's use the quadratic formula:

The quadratic formula states that for an equation of the form $ax^2 + bx + c = 0$, the solutions for $x$ are given by:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In our equation, $a = -2$, $b = -1$, and $c = -1$.

$u = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(-2)(-1)}}{2(-2)}$

$u = \frac{1 \pm \sqrt{1 - 8}}{-4}$

$u = \frac{1 \pm \sqrt{-7}}{-4}$

Since the discriminant $b^2 - 4ac = 1 - 4(-2)(-1) = -7$ is negative, the equation has no real solutions.

Hence, the trigonometric equation $-2\cot^2(x) - \cot(x) - 1 = 0$ has no real solutions.