Logarithmic-trigonomic inequality $\log_5 \sin(x) > \log_{25} \cos(\frac{x}{2}) $ $$\log_5 \sin(x) > \log_{25} \cos(\frac{x}{2})$$
I have simplified it to
$$\sin(\frac{x}{2}) \cdot \sin(\frac{x}{2}) \cdot \cos(\frac{x}{2}) > \frac{1}{4}$$
But what should I do next?
I have no idea about next steps!
Thanks!
 A: $$\log_{5}\sin (x) > \log_{25}\cos (x/2) \implies \log_{5}\sin (x)>\frac 12 \log_{5}\cos (x/2)$$
$$ \implies  \log_{5}(\sin (x))^2 >\log_{5}\cos (x/2) \implies (\sin (x))^2> \cos (x/2)$$
Now use :
$$\sin^2 y =1-\cos ^2 y=1-\left(2\cos^2\left(\frac y2\right)-1\right)^2$$
A: Hint:
Let $t=\cos\frac{x}{2}$. As you pointed out, 
$$\sin^2\frac{x}{2}\cos\frac{x}{2}>\frac{1}{4}$$
$$(1-t^2)t>\frac{1}{4}$$
$$4t^3-4t+1<0$$
Then find the real roots of $4t^3-4t+1$.
A: Note that
$$\log_5 \sin(x) = \frac{\log\sin(x)}{\log 5}$$
and
$$\log_{25} \cos(x/2) = \frac{\log\cos(x/2)}{\log 25}$$
so that your inequality reduces to
$$2\log\sin(x) > \log\cos(x/2).$$
This becomes
$$\log\sin^2(x) > \log\cos(x/2).$$
Since the logarithm is a monotonically increasing function, then
$$\sin^2(x) > \cos(x/2).$$
But we have
$$\sin^2(x) = 1 - \cos^2(x) = 1 - \bigg(2\cos^2(x/2) - 1\bigg)^2 = 1 - 4\cos^4(x/2) + 4\cos^2(x/2) - 1 = -4\cos^4(x/2) + 4\cos^2(x/2).$$
We obtain
$$-4\cos^4(x/2) + 4\cos^2(x/2) > \cos(x/2).$$
Now, set $u = \cos(x/2)$, and get
$$4u^4 - 4u^2 + u < 0$$
$$u(4u^3 - 4u + 1) < 0$$
Using WolframAlpha, we obtain
$$u > 0.837565$$
or
$$0 < u < 0.269594$$
or
$$u < -1.10716.$$
Therefore,
$$\cos(x/2) > 0.837565$$
or
$$0 < \cos(x/2) < 0.269594$$
or
$$\cos(x/2) < -1.10716.$$
The last inequality is extraneous.
Since $\cos(x/2)$ is decreasing, this means that
$$x/2 < \cos^{-1}(0.837565)$$
or
$$\cos^{-1}(0.269594) < x/2 < \cos^{-1}(0)$$
from which we get
$$x/2 < 0.577985$$ 
or
$$1.297825 < x/2 < \frac{\pi}{2}$$
so that the solution set for the original inequality is

$$\{x < 1.15597\} \cup \{2.59565 < x < \pi\}.$$

