Question

4cos2x+4cosx-3=0 2 це степінь

Answer 1

cos x=a, a [-1;1]

4a^2+4a-3=0

   a = -3/2( посторонний)

a=1/2

cos x =1/2

x= +- п/3+2пn

Answer 2

4(соs2x+cosx -3)=0

4(cos3x-3)=0

4(cosx)=0

4=0

cosx=0

x=п/2+пn.где n целое число

 

Answer 3

To solve the equation $4\cos(2x) + 4\cos(x) - 3 = 0$, where $2$ is the exponent, we can use a substitution to simplify it. Let's denote $u = \cos(x)$. Then, we have:

$4u^2 + 4u - 3 = 0.$

Now, we can solve this quadratic equation for $u$. The solutions can be found using the quadratic formula:

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

For our equation $4u^2 + 4u - 3 = 0$, the coefficients are $a = 4$, $b = 4$, and $c = -3$. Substituting these values into the quadratic formula:

$u = \frac{-4 \pm \sqrt{4^2 - 4(4)(-3)}}{2(4)}.$

Simplifying further:

$u = \frac{-4 \pm \sqrt{16 + 48}}{8}.$

$u = \frac{-4 \pm \sqrt{64}}{8}.$

$u = \frac{-4 \pm 8}{8}.$

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

1. $u_1 = \frac{-4 + 8}{8} = \frac{4}{8} = \frac{1}{2},$
2. $u_2 = \frac{-4 - 8}{8} = \frac{-12}{8} = -\frac{3}{2}.$

Remember that $u = \cos(x)$. So, for $u_1 = \frac{1}{2}$, $x$ can be found using the inverse cosine function:

$x_1 = \cos^{-1}\left(\frac{1}{2}\right).$

For $u_2 = -\frac{3}{2}$, there is no real solution, as the cosine function is bound between $-1$ and $1$.

Therefore, the real solution for $x$ is $x_1 = \cos^{-1}\left(\frac{1}{2}\right)$, and any other solutions found in the process that involve complex numbers can be disregarded if you're looking for real solutions.