Question

10 sin^2+11 sin x-8=0

Answer 1

10 sin^2x+11 sin x-8=0    sinx=z
10z²+11z-8=0    √D=√121+320=21
z1=1/20[-11-21]<-1
z2=1/20[-11+21]=1/2
sinx=1/2
x=(-1)^n*π/6+πn  n∈Z

Answer 2

Замена: sinx = t; -1 ≤ t ≤ 1
10t² + 11t - 8 = 0| :10
t² + 1,1t - 0,8 = 0
Согласно теореме Виета
t1 + t2 = -1,1
t1t2 = -0,8
t1 = -1,6 ∅ ; t2 = 0,5
Возвращаемся к замене:
sinx = 0,5
x = (-1)ⁿarcsin0,5 + πn, n є Z;
x = (-1)ⁿπ/6 + πn; n є Z
Ответ: x = (-1)ⁿπ/6 + πn; n є Z

Answer 3

To solve the quadratic equation $10\sin^2 x + 11\sin x - 8 = 0$, we can use the quadratic formula. The general form of a quadratic equation is $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are constants.

In our case: $a = 10$, $b = 11$, $c = -8$.

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

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

Substituting the values from our equation:

$x = \frac{-11 \pm \sqrt{11^2 - 4 \cdot 10 \cdot (-8)}}{2 \cdot 10}$

$x = \frac{-11 \pm \sqrt{121 + 320}}{20}$

$x = \frac{-11 \pm \sqrt{441}}{20}$

$x = \frac{-11 \pm 21}{20}$

Now we have two possible solutions:

1. When $x = \frac{-11 + 21}{20} = \frac{10}{20} = \frac{1}{2}$
2. When $x = \frac{-11 - 21}{20} = \frac{-32}{20} = \frac{-8}{5}$

So, the solutions to the equation are $x = \frac{1}{2}$ and $x = \frac{-8}{5}$.