Question

Cos(x+pi/3)*cos(x-pi/3)=-1/2Хэээээллллппп пллиииз

Answer 1

Есть формула:

cos a*cos b = 1/2*[cos(a-b) + cos(a+b)]

В нашем случае:

cos(x+pi/3)*cos(x-pi/3) = 1/2*[cos(x+pi/3-x+pi/3) + cos(x+pi/3+x-pi/3)] = -1/2

Умножаем все на 2 и приводим подобные в скобках.

cos(2pi/3) + cos(2x) = -1

-1/2 + cos(2x) = -1

cos(2x) = -1/2

2x = +-(2pi/3) + 2pi*k; k € Z

x = +-pi/3 + pi*k; k € Z

Answer 2

It seems like you're looking for a solution to the equation:

$\cos\left(x + \frac{\pi}{3}\right) \cdot \cos\left(x - \frac{\pi}{3}\right) = -\frac{1}{2}$

To solve this equation, we can use the trigonometric identity for the product of two cosines:

$\cos(A) \cdot \cos(B) = \frac{1}{2} \left( \cos(A + B) + \cos(A - B) \right)$

Applying this identity to the given equation:

$\frac{1}{2} \left( \cos\left(x + \frac{\pi}{3} + x - \frac{\pi}{3}\right) + \cos\left(x + \frac{\pi}{3} - x + \frac{\pi}{3}\right) \right) = -\frac{1}{2}$

Simplifying further:

$\frac{1}{2} \left( \cos(2x) + \cos\left(\frac{2\pi}{3}\right) \right) = -\frac{1}{2}$

Now, since $\cos(\frac{2\pi}{3}) = -\frac{1}{2}$, the equation simplifies to:

$\frac{1}{2} \left( \cos(2x) - \frac{1}{2} \right) = -\frac{1}{2}$

Multiply both sides by 2:

$\cos(2x) - \frac{1}{2} = -1$

Add $\frac{1}{2}$ to both sides:

$\cos(2x) = -\frac{1}{2}$

To solve for $x$, you need to find the angles $2x$ for which the cosine is $-\frac{1}{2}$. The angles that satisfy this are $2x = \frac{2\pi}{3} + 2\pi k$ or $2x = \frac{4\pi}{3} + 2\pi k$, where $k$ is an integer.

Dividing both sides by 2:

$x = \frac{\pi}{3} + \pi k$ or $x = \frac{2\pi}{3} + \pi k$

So, the solutions for $x$ are $x = \frac{\pi}{3} + \pi k$ or $x = \frac{2\pi}{3} + \pi k$, where $k$ is an integer. These values of $k$ will give you all possible solutions to the equation.