Question

4log4^2(2sinx)-9log4(2sinx)+2=0

Answer 1

4log4^2(2sinx)-9log4(2sinx)+2=0

log4(2sinx)=z
4z²-9z+2=0   √D=√81-32=7
z1=1/8[9-7]=1/4   2sinx=4^1/4=√2
2sinx=1/8[9+7]=2  2sinx=4²=16  Не подходит.

sinx=√2/2   x=(-1)^n*π/4+πn    n∈Z

Answer 2

To solve the equation 4log₄(2sinx) - 9log₄(2sinx) + 2 = 0, we can use logarithmic properties to simplify it and then solve for sinx. Here's how you can do it step by step:

1. Use logarithmic properties to simplify the equation: 4log₄(2sinx) - 9log₄(2sinx) + 2 = 0

   Combine the logarithms using the properties of logarithms: log₄((2sinx)^4) - log₄((2sinx)^9) + 2 = 0

2. Apply the properties of logarithms to simplify further: log₄((2sinx)^4 / (2sinx)^9) + 2 = 0

3. Combine the terms inside the logarithm using the properties of exponents: log₄((2sinx)^(4 - 9)) + 2 = 0

   log₄((2sinx)^(-5)) + 2 = 0

4. Use the property of logarithms that allows you to bring the exponent down as a coefficient: -5 * log₄(2sinx) + 2 = 0

5. Move the constant term to the other side of the equation: -5 * log₄(2sinx) = -2

6. Divide both sides by -5 to isolate the logarithmic expression: log₄(2sinx) = -2/5

7. Now, we need to remove the logarithm. Rewrite the equation in exponential form: 2sinx = 4^(-2/5)

8. Simplify the right side of the equation: 2sinx = 2^(-1)

9. Divide both sides by 2 to solve for sinx: sinx = 2^(-1) / 2

   sinx = 1/2

So, the solution to the equation 4log₄(2sinx) - 9log₄(2sinx) + 2 = 0 is sinx = 1/2.