Question

2sin^2x/cosx+1=1 Помогите пожалуйста)

Answer 1

2sin²x/(cosx+1)=1,
cosx+1≠0, cosx≠-1
2sin²x=cosx+1,
2*(1-cos²x)-cosx-1=0
2-2cos²x-cosx-1=0
2cos²x+cosx-1=0
cosx=t,   t∈(-1;1]
2t²+t-1=0
t₁=-1,   t₂=1/2
обратная замена:
t=1/2,  cosx=1/2,  x=+-arccos(1/2)+2πn, n∈Z,  x=+-π/3+2πn, n∈Z

Answer 2

Solving the Equation 2sin^2x/cosx+1=1

To solve the equation 2sin^2x/cosx+1=1, we can start by simplifying the equation and then solving for x.

First, let's simplify the equation: 2sin^2x/cosx + 1 = 1 2sin^2x/cosx = 0

Now, let's solve for x using the given information from the search results.

The search results provide various trigonometric equations and identities, but they don't directly address the specific equation 2sin^2x/cosx+1=1. However, we can use the trigonometric identities to solve the equation.

Using the identity sin^2(x) + cos^2(x) = 1, we can rewrite sin^2(x) in terms of cos(x): sin^2(x) = 1 - cos^2(x)

Substituting sin^2(x) = 1 - cos^2(x) into the original equation: 2(1 - cos^2(x))/cos(x) = 0 2 - 2cos^2(x))/cos(x) = 0 2cos(x) - 2cos^3(x) = 0 2cos(x)(1 - cos^2(x)) = 0 2cos(x)sin^2(x) = 0

From this point, we can see that the equation simplifies to 2cos(x)sin^2(x) = 0. This equation has solutions when either cos(x) = 0 or sin(x) = 0.

Therefore, the solutions for x are: - When cos(x) = 0, x = ±π/2 + πk, where k is an integer. - When sin(x) = 0, x = πn, where n is an integer.

So, the solutions for the equation 2sin^2x/cosx+1=1 are x = ±π/2 + πk and x = πn, where k and n are integers.

I hope this helps! If you have further questions or need additional assistance, feel free to ask.