Question

4log4^2(sin^3x)+8log2(sinx)>=1

Answer 1

$4\, log_4^2(sin^3x)+8\, log_2(sinx)\geq 1\; ,\qquad ODZ:\; \; sinx>0\; ,\\\\\star \; \; log_4(sin^3x)=log_{2^2}(sinx)^3=\frac{3}{2}\cdot log_2(sinx)\; \; \star \\\\t=log_2(sinx)\; ,\; \; 4t^2+8t-1\geq 0\; ,\\\\D/4=16+9=25\; ,\; t_1=-1\; ,\; t_2=\frac{1}{9}\\\\4(t+1)(t-\frac{1}{9})\geq 0\; \; ,\quad +++(-1)---(\frac{1}{9})+++\\\\t\leq -1\; \; \; ili\; \; \; t\geq \frac{1}{9}\\\\a)\; \; log_2(sinx)\leq -1\; \; ,\; \; sinx\leq 2^{-1}\; ,\; \; sinx\leq \frac{1}{2}\; ,\; \; 0$

$x\in (2\pi n\, ;\, \frac{\pi }{6}+2\pi n\, ]\cup [\, \frac{5\pi }{6}\, ;\, \pi +2\pi n)\; \; ,\; n\in Z\\\\b)\; \; log_2(sinx)\geq \frac{1}{9}\; \; ,\; \; sinx\geq 2^{1/9}\; \; \; i\; \; \; 01\; \; \Rightarrow \; \; x\in \varnothing \\\\Otvet:\; \; x\in (2\pi n\, ;\, \frac{\pi }{6}+2\pi n\, ]\cup [\, \frac{5\pi }{6}\, ;\, \pi +2\pi n)\; ,\; n\in Z$

Answer 2

To solve the inequality 4log₄(4²(sin³x)) + 8log₂(sin(x)) ≥ 1, we can simplify it step by step.

Let's start with the logarithmic properties. Using the property logₐ(b^c) = c logₐ(b), we can rewrite the inequality as:

4(2log₄(4)(sin³x)) + 8log₂(sin(x)) ≥ 1

Since logₐ(a) = 1 for any base a, we have:

4(2(sin³x)) + 8log₂(sin(x)) ≥ 1

Simplifying further:

8(sin³x) + 8log₂(sin(x)) ≥ 1

Now, let's work on the logarithmic term. Using the change of base formula, we can rewrite log₂(sin(x)) in terms of the natural logarithm (ln):

log₂(sin(x)) = ln(sin(x)) / ln(2)

Substituting this back into the inequality:

8(sin³x) + 8(ln(sin(x)) / ln(2)) ≥ 1

To simplify the inequality further, let's multiply all terms by ln(2) to eliminate the fraction:

8ln(2)(sin³x) + 8(ln(sin(x)) / ln(2)) * ln(2) ≥ 1 * ln(2)

This simplifies to:

8ln(2)(sin³x) + 8ln(sin(x)) ≥ ln(2)

Combining the terms:

8ln(2)(sin³x) + 8ln(sin(x)) ≥ ln(2)

Now, we can divide both sides of the inequality by 8ln(2):

(sin³x) + ln(sin(x)) ≥ ln(2) / (8ln(2))

Simplifying further:

(sin³x) + ln(sin(x)) ≥ 1/8

This is the simplified form of the given inequality.