Question

1+2sinx-cos²x=0.помогите решить,пожалуйста

Answer 1

1+2sinx-cos²x=0
1+2sinx-(1-sin²x) =0
1+2sinx - 1 + sin²x=0
2sinx + sin²x = 0

sinx(2+sinx)=0
 
sinx=0                    sinx  = - 2 , не принадлежит [-1;1]
x=Пk,k э z             
                          

Answer 2

Solving the Equation 1+2sinx-cos²x=0

To solve the equation 1+2sinx-cos²x=0, we can follow these steps:

1. Rearrange the Equation: We start by rearranging the equation to make it easier to solve.

2. Factorize and Solve for x: We then factorize the equation and solve for the variable x.

3. Check for Extraneous Solutions: Finally, we check for extraneous solutions to ensure that the solutions obtained are valid.

Let's go through these steps in detail.

Rearranging the Equation

The given equation is 1+2sinx-cos²x=0. We can rearrange it to make it easier to solve.

Factorize and Solve for x

After rearranging the equation, we can factorize and solve for the variable x.

The equation can be factorized as follows: 2sinx - cos²x = -1

Now, we can solve for x using the factorization: 2sinx - (1 - sin²x) = -1 2sinx - 1 + sin²x = -1 sin²x + 2sinx - 1 = 0

We can solve this quadratic equation for sinx using the quadratic formula: sinx = (-b ± √(b² - 4ac)) / (2a)

Where a = 1, b = 2, and c = -1.

Check for Extraneous Solutions

After obtaining the solutions for sinx, we need to check for extraneous solutions to ensure that the solutions are valid.

The solutions for sinx can then be used to find the corresponding values of x.

Please note that the solutions obtained should be verified for accuracy.

I hope this helps! Let me know if you need further assistance.