Question

8sin^2-2√3cos(π/2-x)-9=0. [-5π/2;-π] 8sin^2-2√3sinx-9=08t^2-2√3t-9=0D=√300t1=2√3+10√3/16=(3√3/4)>1 => пошел далеко и надолгоt2=2√3-10√3/16=(-√3/2)<1 => здравствуйте многоуважаемыйsinx=-√3/2X=-π/3+2πnX=-2π/3+2πnПодходит 1 корень - -7π/3Все правильно?

Answer 1

Ответ:

а)  $x=-2\pi /3+2\pi k\\ x=-\pi /3+2\pi k$, где k∈Z

б) -7π/3

Пошаговое объяснение:

$8sin^2x-2\sqrt{3} sinx-9=0\\ t=sinx, t [-1;1]\\ 8t^2-2\sqrt{3} t-9=0\\ D=12+288=300\\t_1=(2\sqrt{3} +10\sqrt{3} )/16=3\sqrt{3} /4 > 1\\t_2=(2\sqrt{3} -10\sqrt{3} )/16=-\sqrt{3} /2$

$sinx=-\sqrt{3} /2$

$x=-2\pi /3+2\pi k\\ x=-\pi /3+2\pi k$, где k∈Z

Произведём отбор корней.

Рассмотрим первую серию:

-5π/2 ≤ -2π/3+2πk ≤ -π

-11π/6 ≤ 2πk ≤ -π/3

-11/12 ≤ k ≤ -1/6

целых чисел k в этом диапазоне нет

Рассмотрим вторую серию:

-5π/2 ≤ -π/3+2πk ≤ -π

-13π/6 ≤ 2πk ≤ -2π/3

-13/12 ≤ k ≤ -1/3

Подходит целое число k=-1

x=-π/3-2π=-7π/3

Answer 2

Let's go step by step to check your work:

1. You have the equation: 8sin^2x - 2√3cos(π/2-x) - 9 = 0

2. You correctly substituted cos(π/2-x) with sin(x).

3. The equation becomes: 8sin^2x - 2√3sinx - 9 = 0

4. You then attempted to solve this quadratic equation for sin(x) by finding the discriminant (D) and the roots (t1 and t2).

You correctly found the discriminant (D = √300) and the roots as follows:

t1 = (2√3 + 10√3) / 16 = (3√3 / 4) > 1 t2 = (2√3 - 10√3) / 16 = (-√3 / 2) < 1

5. You then attempted to find the values of sin(x) using the roots t1 and t2.

sin(x) = -√3 / 2 (corresponding to t2) - This corresponds to the value of sin(x) when t2 is used.

6. You then found the possible values of x by equating sin(x) to -√3 / 2 and solving for x.

sin(x) = -√3 / 2

x = -π/3 + 2πn or x = -2π/3 + 2πn

7. Finally, you substituted n = 0 to find the specific value of x:

For n = 0: x = -π/3 + 2π * 0 = -π/3

So, one of the correct solutions is x = -π/3.

However, you made a mistake when you substituted n = 0 for the second case:

For n = 0: x = -2π/3 + 2π * 0 = -2π/3

Then, you incorrectly stated that -2π/3 is equal to -7π/3.

The correct solutions for x are: x = -π/3 x = -2π/3

So, in conclusion, you correctly found one of the solutions (x = -π/3), but there is a mistake in your second solution (x = -2π/3).