Đáp án:
1.D
Giải thích các bước giải:
$\cos^4x=(\dfrac{\cos2x+1}{2})^2\\
=\dfrac{1}{4}\left ( \cos^22x+2\cos2x+1 \right )\\
=\dfrac{1}{4}\left ( \dfrac{\cos4x+1}{2}+2\cos2x+1 \right )\\
=\dfrac{1}{4}\left ( \dfrac{\cos4x}{2}+\dfrac{1}{2}+2\cos2x+1 \right )\\
=\dfrac{1}{4}\left ( \dfrac{\cos4x}{2}+\dfrac{3}{2}+2\cos2x \right )\\
=\dfrac{1}{8}\cos4x+\dfrac{3}{8}+\dfrac{1}{2}\cos2x\\
\Rightarrow a=3,b=1,c=1\\
\Rightarrow P=a+b+c=3+1+1=4$
