Question

3cos^2(П/2+x)-2cosx+2cos^2x=0

Answer 1

3cos²( $\pi$|2+x)-2cosx+2cos²x=0
3sin²x-2cosx+2cos²x=0 
3-3cos²x-2cosx+2cos²x=0
cos²x+cosx-3=0
$\left \{ {{y=2} \atop {x=2}} \right.$ cosx=1; cosx=-3 
cosx=-3 (не удовлетворяет условию |cosx| $\leq$1
cosx=1
x=2 $\pi$k, Zэk

Answer 2

To solve the equation 3cos^2(π/2+x) - 2cos(x) + 2cos^2(x) = 0, we can use trigonometric identities to simplify it. Let's go step by step:

First, recall the trigonometric identity: cos^2(x) + sin^2(x) = 1. We can rewrite cos^2(x) as 1 - sin^2(x).

Now, let's substitute cos^2(x) in the equation: 3cos^2(π/2+x) - 2cos(x) + 2cos^2(x) = 0 3(1 - sin^2(π/2+x)) - 2cos(x) + 2(1 - sin^2(x)) = 0

Next, we simplify further: 3 - 3sin^2(π/2+x) - 2cos(x) + 2 - 2sin^2(x) = 0 5 - 3sin^2(π/2+x) - 2cos(x) - 2sin^2(x) = 0

Rearranging the terms: -3sin^2(π/2+x) - 2sin^2(x) - 2cos(x) + 5 = 0

Now, let's substitute sin^2(x) using the identity sin^2(x) = 1 - cos^2(x): -3sin^2(π/2+x) - 2(1 - cos^2(x)) - 2cos(x) + 5 = 0 -3sin^2(π/2+x) - 2 + 2cos^2(x) - 2cos(x) + 5 = 0

Simplifying again: -3sin^2(π/2+x) + 2cos^2(x) - 2cos(x) + 3 = 0

Let's work with the first term -3sin^2(π/2+x): -3sin^2(π/2+x) = -3cos^2(x)

Substituting this back into the equation: -3cos^2(x) + 2cos^2(x) - 2cos(x) + 3 = 0

Combining like terms: -cos^2(x) - 2cos(x) + 3 = 0

To solve this quadratic equation in terms of cos(x), we can factorize it: -(cos(x) - 1)(cos(x) + 3) = 0

Now we have two possible solutions:

1. cos(x) - 1 = 0 => cos(x) = 1
2. cos(x) + 3 = 0 => cos(x) = -3

However, the range of cosine function is -1 to 1, so the second solution, cos(x) = -3, is not valid.

Therefore, the only solution to the equation 3cos^2(π/2+x) - 2cos(x) + 2cos^2(x) = 0 is cos(x) = 1.

Note: In the given equation, x is not explicitly defined, so we cannot determine its exact value from this equation alone.