Question

4 cos^2 x + 11 cos х – 3 =0

Answer 1

Ответ:

Объяснение:

4 cos^2 x + 11 cos х – 3 =0

cosx=t⇒4t²+11t-3=0⇒D=121+48=169; √D=13; t1=(-11-13)/8=-3;

t2=(-11+13)/8=2/8=1/4

cosx=-3⇒x∈∅

cosx=1/4⇒x=±arccos(1/4)+2πk  k∈Z

Answer 2

To solve the quadratic equation $4 \cos^2(x) + 11 \cos(x) - 3 = 0$, we can use the quadratic formula:

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

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

In this case, $a = 4$, $b = 11$, and $c = -3$. Plugging these values into the quadratic formula, we get:

$x = \frac{-11 \pm \sqrt{11^2 - 4 \times 4 \times (-3)}}{2 \times 4}$

Solving the expression inside the square root first:

$11^2 - 4 \times 4 \times (-3) = 121 + 48 = 169$

Now, plugging this value into the quadratic formula:

$x = \frac{-11 \pm \sqrt{169}}{8}$

$x = \frac{-11 \pm 13}{8}$

This gives two potential values for $x$:

1. $x_1 = \frac{-11 + 13}{8} = \frac{1}{4}$
2. $x_2 = \frac{-11 - 13}{8} = -3$

So, the solutions for the equation $4 \cos^2(x) + 11 \cos(x) - 3 = 0$ are $x = \frac{1}{4}$ and $x = -3$.