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B

braga

$$sin^2x+sinxcos4x+cos^24x=\frac{3}{4}$$

[tex]\Leftrightarrow sin^2x+2sinx.\frac{1}{2}cos4x+\frac{1}{4}cos^{24}x-\frac{3}{4}(1-cos^24x)=0 \\ \Leftrightarrow (sinx+\frac{1}{2}cos4x)^2-\frac{3}{4}sin^24x=0 \\ \Leftrightarrow (sinx+\frac{1}{2}cos4x+\frac{\sqrt[]{3}}{2}sin4x)(sinx+\frac{1}{2}cos4x-\frac{\sqrt[]{3}}{2}sin4x)=0 \\ \Leftrightarrow [sinx+sin(4x+\frac{\pi}{6})][sinx+sin(4x-\frac{\pi}{6})]=0[/tex]
 
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