Question

Решите уравнение
${sin}^{2}2x = frac{1}{2}$
​

Answer 1

$sin2x=-frac{sqrt{2} }{2}\\ 2x=(-1)^{k}(-frac{pi}{4}) +pi k, k in Z\\x=(-1)^{k}(-frac{pi}{8}) +frac{pi}{2} k, k in Z$

или

$sin2x=frac{sqrt{2} }{2}\\ 2x=(-1)^{n}(frac{pi}{4}) +pi n, n in Z\\x=(-1)^{n}(frac{pi}{8}) +frac{pi}{2} n, n in Z$

О т в е т.$(-1)^{n}(-frac{pi}{8}) +frac{pi}{2} k, k in Z(-1)^{n}(frac{pi}{8}) +frac{pi}{2} n, n in Z$

Но лучше так:

$sin2x=-frac{sqrt{2} }{2}\\ 2x=-frac{pi}{4} +2pi k, k in Z; 2x=-frac{3pi}{4} +2pi k, k in Z;\\x=-frac{pi}{8} +pi k, k in Z; 2x=-frac{3pi}{8} +pi k, k in Z;$

$sin2x=frac{sqrt{2} }{2}\\ 2x=frac{pi}{4} +2pi n, n in Z;2x=frac{3pi}{4} +2pi n, n in Z;\\x=frac{pi}{8} +pi n, n in Z;x=frac{3pi}{8} +pi n, n in Z;$

О т в е т. $pmfrac{pi}{8} +pi k, k in Z;pmfrac{3pi}{8} +pi n, n in Z;$