• Предмет: Алгебра
  • Автор: MountainDew09
  • Вопрос задан 10 лет назад

√2cos^3x-√2cosx+sin^2x=0

Ответы

Ответ дал: NIXON47
0
 sqrt{2}cos^{3}(x)-sqrt{2}cos(x)+sin^{2}(x)=0;\
sqrt{2}cos(x)(cos^{2}(x) - 1) + sin^{2}(x) = 0;\
-sqrt{2}cos(x)sin^{2}(x) + sin^{2}(x) = 0;\
sin^{2}(x)(1 - sqrt{2}cos(x)) = 0;\
 left[begin{array}{c} sin(x) = 0\cos(x) = frac{sqrt{2}}{2}end{array}right \
left[begin{array}{c} x = left[begin{array}{c} arcsin(0) + 2pi k\pi - arcsin(0) +2pi n end{array}right\
\x = left[begin{array}{c} arccos( frac{sqrt{2}}{2}) + 2pi m\-arccos(frac{sqrt{2}}{2}) + 2pi l end{array}right
end{array}right  k,n,l,m in mathbb Z

Ответ:
left[begin{array}{c} x = left[begin{array}{c} 2pi k\pi +2pi n end{array}right\ \x = left[begin{array}{c} frac{pi}{4} + 2pi m\-frac{pi}{4} + 2pi l end{array}right end{array}right  k,n,l,m in mathbb Z
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