Answered

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sin^22x - 2cos2x + 2=0

Sagot :

evgeniylevi

Answer:

x=πn, n∈Z

Step-by-step explanation:

if sin²2x=1-cos²2x, then

1-cos²2x-2cos2x+2=0; ⇒ cos²2x+2cos2x-3=0; ⇔ (cos2x+3)(cos2x-1)=0;

[tex]\left[\begin{array}{cc}cos2x+3=0\\cos2x-1=0\\\end{array} \ <=> \ \left[\begin{array}{cc}x \ 'does-not-exist'\\cos2x=1\\\end{array} \ <=> \ 2x=2\pi n, n=Z \ <=> x=\pi n, n=Z[/tex]

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