Question

1. √2sin^2x=sinx 2. 2cos^2x+ √3cosx=0 Помогите срочно!!

Answer 1

Equation 1: √2sin^2x = sinx

To solve the equation √2sin^2x = sinx, we can start by squaring both sides of the equation to eliminate the square root:

(√2sin^2x)^2 = (sinx)^2

Simplifying the left side of the equation:

2sin^2x = sin^2x

Next, we can subtract sin^2x from both sides of the equation:

2sin^2x - sin^2x = 0

This simplifies to:

sin^2x = 0

Taking the square root of both sides:

sinx = 0

Therefore, the solution to the equation √2sin^2x = sinx is sinx = 0.

Equation 2: 2cos^2x + √3cosx = 0

To solve the equation 2cos^2x + √3cosx = 0, we can factor out cosx:

cosx(2cosx + √3) = 0

This equation will be true if either cosx = 0 or 2cosx + √3 = 0.

For cosx = 0, the solutions are x = π/2 + nπ, where n is an integer.

For 2cosx + √3 = 0, we can solve for cosx:

2cosx = -√3

cosx = -√3/2

Using the unit circle or a calculator, we find that the solutions for cosx = -√3/2 are x = 5π/6 + 2nπ and x = 7π/6 + 2nπ, where n is an integer.

Therefore, the solutions to the equation 2cos^2x + √3cosx = 0 are x = π/2 + nπ, 5π/6 + 2nπ, and 7π/6 + 2nπ, where n is an integer.

I hope this helps! Let me know if you have any further questions.