Question

4cosx^2x + 4sinx -1 = 0

Answer 1

4cos²x+4sinx-1=0

4(1-sin²X)+4sinx-1=0

-4sin²x+4sinx+3=0

4sin²x-4sinx-3=0

D=16+4·4·3=64,√D=8

SINX=3/2--нет корней, sinx=-1/2⇒x= (-1)^n arcsin(-1/2)+πn,n∈Z

x=(-1)^(n+1)·π/6+πn,n∈Z.

Answer 2

It seems like there might be a typo in the equation you provided. The term "4cosx^2x" is ambiguous and does not seem to be a valid mathematical expression. It's possible that you meant either of the following:

1. $4\cos^2 x + 4\sin x - 1 = 0$ or
2. $4\cos(x^2) + 4\sin x - 1 = 0$

I'll provide solutions for both cases:

Case 1: $4\cos^2 x + 4\sin x - 1 = 0$

To solve this equation for $x$, you can use the trigonometric identity: $\cos^2 x + \sin^2 x = 1$. Rewrite the equation in terms of $\cos^2 x$ and $\sin^2 x$:

$4\cos^2 x + 4\sin x - 1 = 0$
$4(1 - \sin^2 x) + 4\sin x - 1 = 0$

Let $y = \sin x$:

$4(1 - y^2) + 4y - 1 = 0$
$4 - 4y^2 + 4y - 1 = 0$
$3 - 4y^2 + 4y = 0$

Now, we can factor the quadratic equation:

$4y^2 - 4y + 3 = 0$

Using the quadratic formula, $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$y = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 4 \cdot 3}}{2 \cdot 4}$
$y = \frac{4 \pm \sqrt{16 - 48}}{8}$
$y = \frac{4 \pm \sqrt{-32}}{8}$

Since the discriminant ($-32$) is negative, there are no real solutions for $y$, and consequently, no real solutions for $x$ in this case.

Case 2: $4\cos(x^2) + 4\sin x - 1 = 0$

Solving this equation may not be as straightforward due to the combination of a trigonometric function with an argument containing $x^2$. The solutions would likely involve numerical methods or graphical analysis. However, it's essential to confirm if this is the correct equation you intended to write.

If you have a different equation or if there is a specific range for $x$ that you want to find solutions in, please provide additional information, and I'd be happy to assist further.