Question

Sin^6x+cos^6x=1\4 помогите решить

Answer 1

Ответ:

$sin^6x+cos^6x=\dfrac{1}{4}\\\\(sin^2x)^3+(cos^2x)^3=\dfrac{1}{4}\\\\(\underbrace{sin^2x+cos^2x}_{1})(sin^4x-sin^2x\cdot cos^2x+cos^4x)=\dfrac{1}{4}\\\\(sin^4x+2sin^2x\cdot cos^2x+cos^4x)-2sin^2x\cdot cos^2x-sin^2x\cdot cos^2x=\dfrac{1}{4}\\\\\underbrace{(sin^2x+cos^2x)^2}_{1^2=1}-3sin^2x\cdot cos^2x=\dfrac{1}{4}\\\\3sin^2x\cdot cos^2x=1-\dfrac{1}{4}\ \ ,\ \ \ 3sin^2x\cdot cos^2x=\dfrac{3}{4}\ \ ,\ \ \ sin^2x\cdot cos^2x=\dfrac{1}{4}\ \ ,$

$(sinx\cdot cosx)^2=\dfrac{1}{4}\ \ ,\ \ \ \Big(\dfrac{1}{2}\cdot sin2x\Big)^2=\dfrac{1}{4}\ \ \ \Rightarrow \ \ \ sin^22x=1\ \ ,\\\\\dfrac{1-cos4x}{2}=1\ \ ,\ \ \ 1-cos4x=2\ \ ,\ \ \ cos4x=-1\ \ ,\\\\4x=\pi +2\pi n\ \ ,\ \ n=0,\pm 1,\pm 2\, ,...\\\\\boxed{\ x=\dfrac{\pi }{4}+\dfrac{\pi n}{2}\ \ ,\ \ n=0,\pm 1,\pm 2,\, ...\ }$