Question

4sin^2x+sinxcosx-3=0

Answer 1

Решение дано на фото.

Answer 2

To solve the quadratic equation 4sin^2x + sinxcosx - 3 = 0, we can use substitution and quadratic formula. Let's proceed step by step:

1. Make a substitution: Let's set u = sin(x). This implies that du/dx = cos(x), and we can rewrite the equation as follows: 4u^2 + u * du - 3 = 0

2. Solve the quadratic equation for u: To solve this quadratic equation for u, we can use the quadratic formula: u = (-B ± √(B^2 - 4AC)) / 2A, where A = 4, B = 1, and C = -3. Plugging in the values, we get: u = (-1 ± √(1 + 48)) / 8 u = (-1 ± √49) / 8 u = (-1 ± 7) / 8

   So, we have two possible solutions for u: u = 1 and u = -3/4

3. Solve for x using the original substitution: Now that we have the values of u, we can solve for x using the original substitution u = sin(x): For u = 1: sin(x) = 1 x = π/2 + 2πn, where n is an integer.

   For u = -3/4: sin(x) = -3/4 Since the sine function can't be larger than 1 or smaller than -1, there are no real solutions for this case.

Therefore, the solution for the equation 4sin^2x + sinxcosx - 3 = 0 is x = π/2 + 2πn, where n is an integer.