Question

3sin^22x+7cos2x-3=0
Помогите

Answer 1

$3sin^22x+7cos2x-3=0\\ 3(1-cos^22x)+7cos2x-3=0 \\ -3cos^22x+7cos2x=0\\ cos2x(-3cos2x+7)=0\\ cos2x=0 \\ 2x= \frac{ \pi }{2}+ \pi n\\x= \frac{ \pi }{4}+ \frac{ \pi }{2} n\\ cos2x= \frac{7}{3}\\2x=+-arccos \frac{7}{3} +2 \pi n$