How do you solve 2cos^2x+3sinx=32cos2x+3sinx=3 in the interval 0<=x<=2pi0≤x≤2π?

1 Answer
Binayaka C.
Sep 26, 2016

Solution x= pi/6,pi/2, (5pi)/6x=π6,π2,5π6 for 0 <=x<=2pi0≤x≤2π

Explanation:

2cos^2x+3sinx=3 or 2(1-sin^2x)+3sinx-3=0 or 2sin^2x-3sinx+1=0 or (2sinx-1)(sinx-1)=02cos2x+3sinx=3or2(1−sin2x)+3sinx−3=0or2sin2x−3sinx+1=0or(2sinx−1)(sinx−1)=0. So either 2sinx-1=0 or sinx-1=0 2sinx−1=0orsinx−1=0 When 2sinx-1=0 :.2sinx=1 :. sinx= 1/2; sin(pi/6)=1/2 and sin(pi-pi/6)=1/2 :. x=pi/6,(5pi)/6 When sinx-1=0 or sinx=1 ; sin (pi/2)=1 :. x= pi/2
Solution x= pi/6,pi/2, (5pi)/6 for 0 <=x<=2pi [Ans]

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