Question

Розв'яжіть рівняння: 1)2sin3x=2 2)3tg²2x-1=0

Answer 1

№1

2sin3x = 2

sin3x = 2 : 2

sin3x = 1

3x = π/2 + 2πn, где n Є Z

x = π/3 + 2πn/3

№2

3tg²2x - 1 = 0

tg²2x = 1

tg2x = 1

2x = π/4 + 2πn

x = π/8 + 2πn/8

x = π/8 + πn/4

Answer 2

1. 2sin(3x) = 2

To solve this equation for x, we can first divide both sides by 2 to simplify it:

sin(3x) = 1

Now, we need to find the values of x for which sin(3x) is equal to 1. Recall that the sine function has a maximum value of 1, and it occurs when the angle is 90 degrees (π/2 radians) or 270 degrees (3π/2 radians). Therefore:

3x = π/2 + 2πn (for the first solution) 3x = 3π/2 + 2πn (for the second solution)

Where n is an integer. Now, we can solve for x:

For the first solution: 3x = π/2 + 2πn x = (π/2 + 2πn) / 3

For the second solution: 3x = 3π/2 + 2πn x = (3π/2 + 2πn) / 3

So, the solutions for the equation 2sin(3x) = 2 are: x = (π/6 + 2πn/3) x = (π/2 + 2πn/3)

2. 3tan²(2x) - 1 = 0

To solve this equation for x, first, add 1 to both sides:

3tan²(2x) = 1

Now, divide both sides by 3:

tan²(2x) = 1/3

To solve for x, we need to take the square root of both sides:

tan(2x) = ±√(1/3)

Now, we need to find the values of 2x for which the tangent is equal to ±√(1/3). The tangent function is positive in the first and third quadrants, so we need to consider both positive and negative square root values:

For positive square root: 2x = arctan(√(1/3))

For negative square root: 2x = -arctan(√(1/3))

Now, solve for x:

For positive square root: x = (1/2) * arctan(√(1/3))

For negative square root: x = -(1/2) * arctan(√(1/3))

These are the solutions for the equation 3tan²(2x) - 1 = 0.