Question

1-cos^2x+4sinx=0 РЕШИТЕ ПОЖАЛУЙСТА :*

Answer 1

${cos=-\sqrt(((4*sinx)/(x))+(1/(x))), cos=\sqrt(((4*sinx)/(x))+(1/(x)))}$

$1-1-sin^2x+4sinx=0$

$sin^2x-4sinx=0$

$sinx_1=0$

$sinx_2=4$

 

 

На график еще

Answer 2

1-1-sin^2x+4sinx=0

sin^2x-4sinx=0

sinx=t

sinx=4 и sinx=0

 

делай выборку, не знаю какой промежуток.

Answer 3

To solve the equation 1 - cos^2(x) + 4sin(x) = 0, we can use trigonometric identities and algebraic manipulation. Let's break it down step by step:

Step 1: Simplify the equation

We can start by using the Pythagorean identity sin^2(x) + cos^2(x) = 1 to simplify the equation. By substituting 1 - cos^2(x) with sin^2(x), we get:

sin^2(x) + 4sin(x) = 0

Step 2: Factor out sin(x)

Next, we can factor out sin(x) from the equation:

sin(x)(sin(x) + 4) = 0

Now we have two factors: sin(x) = 0 and sin(x) + 4 = 0.

Step 3: Solve for sin(x) = 0

For sin(x) = 0, the solutions are x = 0, π, 2π, 3π, and so on. This is because sin(x) = 0 when x is an integer multiple of π.

Step 4: Solve for sin(x) + 4 = 0

For sin(x) + 4 = 0, we can subtract 4 from both sides of the equation:

sin(x) = -4

However, the sine function only takes values between -1 and 1, so there are no real solutions for this equation.

Final Answer:

The solutions to the equation 1 - cos^2(x) + 4sin(x) = 0 are x = 0, π, 2π, 3π, and so on. There are no real solutions for sin(x) + 4 = 0.

Please let me know if you need any further clarification or assistance!