[tex]\displaystyle \text{log}_{\ \displaystyle 2} \ \text{sen x}=\text{log}_{\displaystyle 4} \ \text{cos x} \\\\ \underline{\text{fazendo mudan{\c c}a de base na direita}}: \\\\ \text{log}_{\displaystyle \ \text a}\ \text b = \frac{\text{log}_{\displaystyle \ \text c}\ \text b}{\text{log}_{\displaystyle \ \text c} \ \text a } \\\\\\ \underline{\text{fazendo c = 2}:}\\\\ \text{log}_{\displaystyle \ \text 2}\ \text{sen x} = \frac{\text{log}_{\displaystyle \ 2} \ \text{cos x}}{\text{log}_{\displaystyle \ 2 }\ 4}[/tex]
[tex]\displaystyle \text{log}_{\displaystyle \ 2 }\ \text{sen x }= \frac{1}{2}.\text{log}_{\displaystyle \ 2} \ \text{cos x} \\\\\\ \text{log}_{\displaystyle \ 2}\ \text{sen x } = \text{log}_{\displaystyle \ 2} \ \text{(cos x)}^{\frac{1}{2}} \\\\\\ \text{sen x }= \sqrt{\text{cos(x)}}[/tex]
Se 0º < x < 90º então sen x > 0 e cos x > 0, então não precisamos nos preocupar com módulo.
Elevando ao quadrado dos dois lados :
[tex]\displaystyle \text {sen}^2 \ \text x= \text{cos x} \\\\ 1-\text{cos}^2 \ \text x=\text{cos x} \\\\ \text{cos}^2 \ \text x+\text{cos x}-1 = 0 \\\\ \text{cos x}=\frac{-1+\sqrt{1-4.(-1)}}{2} \\\\\ \boxed{\text{cos x} =\frac{\sqrt{5}-1}{2} \ \approx \ 0,61}[/tex]
Uma análise rápida :
[tex]\displaystyle \text{cos 45}^\circ = \frac{\sqrt 2}{2} \approx 0,7 \ \ \ ; \ \ \ \text{cos 60}^\circ = \frac{1}{2} = 0,5[/tex]
Portanto :
[tex]\text{cos 45}^\circ > \text{cos x}> \text{cos 60}^\circ[/tex]
[tex]\huge\boxed{45^\circ < \text x < 60^\circ\ }\checkmark[/tex]
letra A
