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Question Number 59389 by hovea cw last updated on 09/May/19
If0⩽x⩽πand81sin2x+81cos2x=30,thenxisequalto
Answered by ajfour last updated on 09/May/19
t+81t=30⇒t2−30t+81=0⇒t=3,2781sin2x=8114,8134⇒x=π6,π3,2π3,5π6.
Answered by tanmay last updated on 09/May/19
(81)sin2x+(81)1−sin2x=30(81)sin2x+81(81)sin2x=30(9)2sin2x+92(9)2sin2x=30(9sin2x−99sin2x)2+2×9sin2x×99sin2x=30(9sin2x−99sin2x)2=(23)2(k−9k)2−(23)2=0k−9k+23=0k2+23k−9=0k2+(33−3)k−3×3×3=0k(k+33)−3(k+33)=0(k+33)(k−3)=0k+33≠0k=39sin2x=31232sin2x=312sin2x=14sinx=12=sin(π6)→x=π6(firstquadrant)and(π−π6)i,e5π6insecondquadrantsinx≠−12whenπ⩾x⩾0againk−9k−23=0k2−9−23k=0k2−23k+3−12=0(k−3)2−(23)2=0(k−3+23)(k−3−23)=0k+3≠0k=339sin2x=33232sin2x=332sin2x=34sinx=32→x=π3and(π−π3)i,e2π3hencex=π6,π3,2π3,5π6sinceπ⩾x⩾0sosinx≠negetivesosinx≠−32
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