Question

2sin^2(2x)+7sin(2x)-4=0

Answer 1

По т Виета
sin2x=-4 не удовлетворяет
sin2x=-1/2
x=(-1)^(n+1)*пи/12+пи*k/2,k целое

Answer 2

To solve the equation 2sin²(2x) + 7sin(2x) - 4 = 0, we can use a substitution to simplify the expression. Let's replace sin(2x) with a variable, such as u. The equation becomes:

2u² + 7u - 4 = 0

Now we can solve this quadratic equation for u. We can use the quadratic formula:

u = (-b ± √(b² - 4ac)) / (2a)

In our case, a = 2, b = 7, and c = -4. Substituting these values into the quadratic formula, we get:

u = (-7 ± √(7² - 4 * 2 * -4)) / (2 * 2) u = (-7 ± √(49 + 32)) / 4 u = (-7 ± √81) / 4 u = (-7 ± 9) / 4

This gives us two possible solutions for u:

1. u = (-7 + 9) / 4 = 2 / 4 = 1/2
2. u = (-7 - 9) / 4 = -16 / 4 = -4

Now we substitute back sin(2x) for u:

1. sin(2x) = 1/2
2. sin(2x) = -4

To find the values of x, we need to solve for 2x in each equation and then find x.

1. sin(2x) = 1/2 We know that sin(30°) = 1/2. So, 2x = 30° + k * 360° or 2x = 150° + k * 360°, where k is an integer.

2x = 30° + k * 360° x = (30° + k * 360°) / 2 x = 15° + k * 180°

2x = 150° + k * 360° x = (150° + k * 360°) / 2 x = 75° + k * 180°

2. sin(2x) = -4 There is no solution for this equation because the sine function can only take values between -1 and 1.

Therefore, the solutions for the equation 2sin²(2x) + 7sin(2x) - 4 = 0 are:

x = 15° + k * 180° x = 75° + k * 180°

where k is an integer.