Question

Logx 3x-1/x^2+1>0 и 8cos^4x-cos4x=1

Answer 1

1) logx (3x-1)/>0будем решать крест на крест знаменатель равна 0
x^2+1. logx(3x-1)>0
3x-1>x^0
3x-1>1;;3x>2::::x>2/3

Answer 2

To solve the inequality log(x)(3x - 1)/(x^2 + 1) > 0, we need to analyze the sign of the expression for different intervals of x. Let's break it down step by step:

1. Start by finding the critical points where the numerator and denominator become zero.

   Numerator: 3x - 1 = 0 x = 1/3

   Denominator: x^2 + 1 = 0 (No real solutions since x^2 is always positive)

   Therefore, the critical point is x = 1/3.

2. Analyze the sign of the expression for three intervals: (-∞, 1/3), (1/3, ∞), and x = 1/3.

   Interval (-∞, 1/3): Choose a test point, x = 0, and substitute it into the expression: log(0)(3(0) - 1)/(0^2 + 1) = log(0)(-1) = undefined (log(0) is undefined)

   So, the expression is undefined in this interval.

   Interval (1/3, ∞): Choose a test point, x = 1, and substitute it into the expression: log(1)(3(1) - 1)/(1^2 + 1) = log(1)(2/2) = log(1)(1) = 0

   The expression is positive in this interval.

   x = 1/3: Substitute x = 1/3 into the expression: log(1/3)(3(1/3) - 1)/((1/3)^2 + 1) = log(1/3)(1 - 1)/(1/9 + 1) = log(1/3)(0)/(10/9) = 0

   The expression is zero at x = 1/3.

3. Based on the analysis, we can conclude the solution:

   The expression is undefined for x < 1/3. The expression is zero at x = 1/3. The expression is positive for x > 1/3.

   Therefore, the solution to the inequality is x > 1/3.

Now, let's move on to the equation 8cos^4(x) - cos(4x) = 1:

1. Start by simplifying the equation using the identity cos(4x) = 2cos^2(2x) - 1:

   8cos^4(x) - (2cos^2(2x) - 1) = 1

2. Rearrange the equation:

   8cos^4(x) - 2cos^2(2x) + 1 = 1

3. Combine like terms:

   8cos^4(x) - 2cos^2(2x) = 0

4. Factor out cos^2(x):

   cos^2(x)(8cos^2(x) - 2cos^2(2x)) = 0

5. Apply the zero-product property:

   cos^2(x) = 0 or 8cos^2(x) - 2cos^2(2x) = 0

6. Solve each equation separately:

   For cos^2(x) = 0: cos(x) = 0 x = π/2 + nπ where n is an integer

   For 8cos^2(x) -