Question
3sin 2A +4 cos2A max and min value and solve
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3(sin2θ+cos2θ)+cos2θ=3(1)+cos2θ
=3+(1+cos2θ)/2
=3+1/2+!/2cos2θ
=7/2+1/2cos2θ
min=7/2-√(1/2)2+02
=7/2-1/2=3
max=7/2+√(1/2)2+02
=7/2+1/2=4
C NARENDRA CHARY 8 year ago
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