[toán11]phương trình lượng giác

L

l94

cos2x+2cos22x+2cos23x−3=cos4x(2sin2x+1)cos^2x+2cos^22x+2cos^23x-3=cos^4x(2sin2x+1)
⇔cos2x+cos4x+cos6x=cos4x(2sin2x+1) \Leftrightarrow cos2x+cos4x+cos6x=cos^4x(2sin2x+1)
⇔cos4x(2cos2x+1)=cos4x(2sin2x+1) \Leftrightarrow cos4x(2cos2x+1)=cos^4x(2sin2x+1)
⇔(1−2sin2x)(1+2sin2x)(2cos2x+1)=cos4x(2sin2x+1) \Leftrightarrow (1-2sin2x)(1+2sin2x)(2cos2x+1)=cos^4x(2sin2x+1)
⇔2sin2x+1=0 \Leftrightarrow 2sin2x+1=0
và (1−2sin2x)(2cos2x+1)=cos4x (1-2sin2x)(2cos2x+1)=cos^4x
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