pt luong giac

N

nguyenbahiep1

3 4(sin^4 x+ cos^4 x) + \sqrt[2]{3}sin4x= 2

câu 3

[laTEX]4(1 - 2sin^2x.cos^2x) + \sqrt{3}.sin4x - 2 = 0 \\ \\ 4 - 2sin^22x + \sqrt{3}.sin4x - 2 = 0 \\ \\ \sqrt{3}.sin4x - 1 + cos4x +2 = 0 \\ \\ \sqrt{3}.sin4x + cos4x = - 1 \\ \\ 2 sin(4x + \frac{\pi}{6}) = - 1 [/laTEX]
 
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