Question

CРОООЧНО!! 2cos²x+5sinx-4=0

Answer 1

Держи решение...........

Answer 2

To solve the trigonometric equation 2cos²x + 5sinx - 4 = 0, we can use some trigonometric identities to simplify the equation and find its solutions.

Let's start by using the identity: cos²x + sin²x = 1 Rearrange the equation: 2cos²x + 5sinx - 4 = 0 2cos²x - 4 + 5sinx = 0 2(cos²x - 2) + 5sinx = 0

Now, we can use the identity: 1 - sin²x = cos²x Substitute the identity into the equation: 2(1 - sin²x) - 4 + 5sinx = 0 2 - 2sin²x - 4 + 5sinx = 0 -2sin²x + 5sinx - 2 = 0

Now, we have a quadratic equation in terms of sinx: -2sin²x + 5sinx - 2 = 0

To solve this quadratic equation, we can factor it: (-2sinx + 1)(sinx - 2) = 0

Now, set each factor to zero and solve for sinx:

1. -2sinx + 1 = 0 -2sinx = -1 sinx = 1/2

2. sinx - 2 = 0 sinx = 2 (But this is not possible since the sine function's range is between -1 and 1)

So, the only valid solution for the given equation is sinx = 1/2.

Now, we can find the corresponding values of x using inverse trigonometric functions: x = arcsin(1/2)

The principal value of arcsin(1/2) is π/6, so the general solution for x is: x = π/6 + 2πn, where n is an integer.

So, the final solutions for the equation 2cos²x + 5sinx - 4 = 0 are: x = π/6 + 2πn, where n is an integer.