Question

3 cos^2 x + 7 sin x- 5=0

Answer 1

3-3sin²x+7sinx-5=0
sinx=a
3a²-7a+2=0
D=49-24=25
a1=(7-5)/6=1/3⇒sinx=1/3⇒x=(-1)^n*arcsin1/3+πk,k∈z
a2=(7+5)/6=2⇒sinx=2>1 нет решения

Answer 2

To solve the equation $3 \cos^2 x + 7 \sin x - 5 = 0$, we can use the trigonometric identity $\cos^2 x = 1 - \sin^2 x$. Let's first substitute this identity into the equation:

$3(1 - \sin^2 x) + 7 \sin x - 5 = 0$

Now, let's simplify the equation:

$3 - 3\sin^2 x + 7\sin x - 5 = 0$

Rearrange the terms:

$-3\sin^2 x + 7\sin x - 2 = 0$

To solve this quadratic equation, we can use the quadratic formula:

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

Where in our equation, $a = -3$, $b = 7$, and $c = -2$.

Now, plug in these values into the quadratic formula:

$x = \frac{-7 \pm \sqrt{7^2 - 4(-3)(-2)}}{2(-3)}$

$x = \frac{-7 \pm \sqrt{49 - 24}}{-6}$

$x = \frac{-7 \pm \sqrt{25}}{-6}$

$x = \frac{-7 \pm 5}{-6}$

Now, we have two possible solutions for $x$:

1. $x = \frac{-7 + 5}{-6} = \frac{-2}{-6} = \frac{1}{3}$

2. $x = \frac{-7 - 5}{-6} = \frac{-12}{-6} = 2$

So the solutions to the equation are $x = \frac{1}{3}$ and $x = 2$.