[toán 11] lượng giác

B

bibau179

ta có {sin(x-pi/4)}^{2}={cosx}^{2}\Rightarrow\frac{1-cos(2x-pi/2)}{2}=\frac{1+cos2x}{2}\Leftrightarrowsin2x+cos2x=0\Leftrightarrow \sqrt[2]{2}cos(2x-pi/4)=0=>2x-pi/4=pi/2+kpi=>x=3pi/8+k pi/2
 
B

bibau179

sin4x+2cos2x=0\Leftrightarrow 2sin2xcos2x+2cos2x=0 \Leftrightarrow 2cos2x(sin2x+1)=0 \Leftrightarrow cos2x=0 hoặc sin2x=-1
với cos2x=0 ta có nghiệm x=pi/4+kpi/2
với sin2x=-1=>x=-pi/4+kpi
 
N

newtons007

a) [TEX]sin^2(x-\frac{pi}{4})= cos^2x \Leftrightarrow 1-cos(2x-\frac{pi}{2})=2cos^2x \Leftrightarrow 1- sin(2x)=2cos^2x \Leftrightarrow 2cos^2x+sin2x-1=0 [/TEX] chia 2 vế cho [TEX]cos^2x \Leftrightarrow 2 + 2tanx-(1+tan^2x)=0 \Leftrightarrow tan^2x -2tanx -1 =0[/TEX] ti nua giai tiep ko piet lam dấu [TEX][[/TEX]làm sao giai cau 2 dây hai`..zz
 
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