Question

2 sinx*cosx+√3 cosx = 2sinx+√3

Answer 1

Ответ:

Пошаговое объяснение:

cosx(2sinx+V3)-(2sinx+V3)=0,  (2sinx+V3)(cosx-1)=0,   2sinx+V3=0  или  cosx-1=0,    sinx=-V3/2,   x=(-1)^n* (-п/3)+пк,  cosx=1,  x=2пк,   к  E   Z

Answer 2

To solve the equation 2sin(x)cos(x) + √3cos(x) = 2sin(x) + √3, we can first simplify the equation and then solve for x.

Let's begin:

2sin(x)cos(x) + √3cos(x) = 2sin(x) + √3

First, we notice that both terms on the left side of the equation contain cos(x). We can factor out cos(x) to simplify the equation:

cos(x)(2sin(x) + √3) = 2sin(x) + √3

Now, we have two cases to consider:

Case 1: cos(x) = 0

If cos(x) = 0, then the left side of the equation becomes zero. We can substitute cos(x) = 0 into the original equation to check if it satisfies the equation:

2sin(x)(0) + √3(0) = 2sin(x) + √3 0 = 2sin(x) + √3

As there is no value of sin(x) that can satisfy this equation, there are no solutions in this case.

Case 2: (2sin(x) + √3) ≠ 0

In this case, we can divide both sides of the equation by (2sin(x) + √3):

cos(x) = (2sin(x) + √3) / (2sin(x) + √3)

Simplifying further, we have:

cos(x) = 1

Now, we know that the range of the cosine function is [-1, 1]. Since cos(x) = 1, we can conclude that x must be an angle whose cosine is equal to 1. Therefore, x = 2πn, where n is an integer.

In summary, the equation 2sin(x)cos(x) + √3cos(x) = 2sin(x) + √3 has no solutions when cos(x) = 0, and when (2sin(x) + √3) ≠ 0, the solution is x = 2πn, where n is an integer.