Question

4x^2+x-3=0; √x+2=3-√x-1; 7^=49^x+2; Cos 2x = 1/2 помогите решить пожалуйста

Answer 1

Смотри вложения......................................

Answer 2

Solving the equations:

The equations given are:

1) 4x^2 + x - 3 = 0 2) √x + 2 = 3 - √x - 1 3) 7^x = 49^x + 2 4) Cos 2x = 1/2

Let's solve these one by one:

1) 4x^2 + x - 3 = 0 Using the quadratic formula: D = b^2 - 4ac = 1^2 - 4*4*-3 = 17 x = (-b ± √D) / (2a) x = (-1 ± √17) / 8 x = 1, -3/4

2) √x + 2 = 3 - √x - 1 Squaring both sides: (√x + 2)2 = (3 - √x - 1)2 x + 4√x + 4 = 9 - 6√x + x + 2√x + 1 4√x = 5 √x = 5/4 x = 25/16

3) 7^x = 49^x + 2 Taking logarithms (base 7) of both sides: x = 2x + log7(2) x = 2 x = 2

4) Cos 2x = 1/2 Using the trig identity: Cos 2x = 1 - 2Sin^2 x 1/2 = 1 - 2Sin^2 x Sin^2 x = 1/4 x = π/6 or 5π/6