Решить уравнение:[tex] \sqrt{2} cos(2x+ \frac{pi}{4} )+1=0[/tex][tex]2cos(

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$\sqrt{2} cos(2x+ \frac{\pi}{4} )+1=0 \\\ cos(2x+ \frac{\pi}{4} )=- \frac{1}{ \sqrt{2} } = -\frac{\sqrt{2}}{ 2 } \\\ \left \{ {{2x+\frac{\pi}{4}= \frac{2\pi}{3}+2\pi n } \atop {2x+\frac{\pi}{4}=- \frac{2\pi}{3}+2\pi n}} ight. \\\ \left \{ {{2x= \frac{5\pi}{12}+2\pi n } \atop {2x=- \frac{11\pi}{12}+2\pi n}} ight. \\\ \left \{ {{x= \frac{5\pi}{24}+\pi n } \atop {x=- \frac{11\pi}{24}+\pi n}} ight., n\in Z$ $2cos( \frac{\pi}{3} - 3x) - \sqrt{3} =0 \\\ cos( \frac{\pi}{3} - 3x)= \frac{\sqrt{3}}{2} \\\ \left \{ {{\frac{\pi}{3} - 3x= \frac{\pi}{6}+2\pi n } \atop {\frac{\pi}{3} - 3x=- \frac{\pi}{6}+2\pi n}} ight. \\\ \left \{ {{3x= \frac{\pi}{6}-2\pi n } \atop { 3x=- \frac{\pi}{2}-2\pi n}} ight. \\\ \left \{ {{x= \frac{\pi}{18}- \frac{2\pi n}{3} } \atop { x=- \frac{\pi}{6}-\frac{2\pi n}{3}}} ight., n\in Z$ $5sin(2x-1)-2=0 \\\ sin(2x-1)= \frac{2}{5} \\\ 2x-1=(-1)^k arcsin \frac{2}{5} +\pi k \\\ 2x=(-1)^k arcsin \frac{2}{5}+1 +\pi k \\\ x= \frac{(-1)^k arcsin \frac{2}{5}}{2}+ \frac{1}{2} + \frac{\pi k}{2}, k\in Z$
- Автор:
  giggles30st
- 6 лет назад
- 0
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