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Câu hỏi: Tổng các nghiệm của phương trình ${{\sin }^{4}}x+{{\sin }^{4}}\left( x+\dfrac{\pi }{4} \right)+{{\sin }^{4}}\left( x-\dfrac{\pi }{4} \right)=\dfrac{5}{4}$ trong khoảng $\left( 0;\pi \right)$ bằng?
A. $2\pi .$
B. $4\pi .$
C. $\pi .$
D. $3\pi .$
A. $2\pi .$
B. $4\pi .$
C. $\pi .$
D. $3\pi .$
Ta có: ${{\sin }^{4}}\left( x+\dfrac{\pi }{4} \right)+{{\sin }^{4}}\left( x-\dfrac{\pi }{4} \right)={{\left[ \dfrac{1}{\sqrt{2}}\left( \sin x+\cos x \right) \right]}^{4}}+{{\left[ \dfrac{1}{\sqrt{2}}\left( \sin x-\cos x \right) \right]}^{4}}$
$=\dfrac{1}{4}{{\left( \sin x+\cos x \right)}^{4}}+\dfrac{1}{4}{{\left( \sin x-\cos x \right)}^{4}}$
$=\dfrac{1}{4}{{\left[ {{\left( \sin x+\cos x \right)}^{2}}+{{\left( \sin x-\cos x \right)}^{2}} \right]}^{2}}-\dfrac{1}{2}{{\left( \sin x+\cos x \right)}^{2}}{{\left( \sin x-\cos x \right)}^{2}}.$
$=1-\dfrac{1}{2}{{\left( {{\sin }^{2}}x-{{\cos }^{2}}x \right)}^{2}}=1-\dfrac{1}{2}{{\cos }^{2}}2x.$
${{\sin }^{4}}x+{{\sin }^{4}}\left( x+\dfrac{\pi }{4} \right)+{{\sin }^{4}}\left( x-\dfrac{\pi }{4} \right)=\dfrac{5}{4}\Leftrightarrow {{\left( \dfrac{1-\cos 2x}{2} \right)}^{2}}+1-\dfrac{1}{2}{{\cos }^{2}}2x=\dfrac{5}{4}.$
$\Leftrightarrow {{\cos }^{2}}2x+2\cos 2x=0\Leftrightarrow \cos 2x=0\Leftrightarrow x=\dfrac{\pi }{4}+k\dfrac{\pi }{2}\left( k\in \mathbb{Z} \right).$
Mà $x\in \left( 0;\pi \right)\Rightarrow x=\dfrac{\pi }{4};x=\dfrac{3\pi }{4}.$ Tổng các nghiệm là $S=\dfrac{\pi }{4}+\dfrac{3\pi }{4}=\pi .$
$=\dfrac{1}{4}{{\left( \sin x+\cos x \right)}^{4}}+\dfrac{1}{4}{{\left( \sin x-\cos x \right)}^{4}}$
$=\dfrac{1}{4}{{\left[ {{\left( \sin x+\cos x \right)}^{2}}+{{\left( \sin x-\cos x \right)}^{2}} \right]}^{2}}-\dfrac{1}{2}{{\left( \sin x+\cos x \right)}^{2}}{{\left( \sin x-\cos x \right)}^{2}}.$
$=1-\dfrac{1}{2}{{\left( {{\sin }^{2}}x-{{\cos }^{2}}x \right)}^{2}}=1-\dfrac{1}{2}{{\cos }^{2}}2x.$
${{\sin }^{4}}x+{{\sin }^{4}}\left( x+\dfrac{\pi }{4} \right)+{{\sin }^{4}}\left( x-\dfrac{\pi }{4} \right)=\dfrac{5}{4}\Leftrightarrow {{\left( \dfrac{1-\cos 2x}{2} \right)}^{2}}+1-\dfrac{1}{2}{{\cos }^{2}}2x=\dfrac{5}{4}.$
$\Leftrightarrow {{\cos }^{2}}2x+2\cos 2x=0\Leftrightarrow \cos 2x=0\Leftrightarrow x=\dfrac{\pi }{4}+k\dfrac{\pi }{2}\left( k\in \mathbb{Z} \right).$
Mà $x\in \left( 0;\pi \right)\Rightarrow x=\dfrac{\pi }{4};x=\dfrac{3\pi }{4}.$ Tổng các nghiệm là $S=\dfrac{\pi }{4}+\dfrac{3\pi }{4}=\pi .$
Đáp án C.