How do you solve cos^2x=1-sinx between the interval 0<=x<=2pi?

1 Answer
Binayaka C.
Sep 29, 2016

Solution; x=0,x=pi/2,x=pi;x=2pi. In interval 0<=x<=2pi

Explanation:

cos^2x=1-sinx or 1-sin^2x= 1-sinx or sin^2x-sinx=0 or sinx(sinx-1)=0 :. either sinx=0 or sinx-1=0 :. sinx=1 When sinx=0; x=0,x=pi,x=2pi. Since sin0=0; sin pi=0 ; sin2pi=0
When sinx=1; x=pi/2. Since sin (pi/2)=0
Solution; x=0,x=pi/2,x=pi;x=2pi. In interval 0<=x<=2pi[Ans]

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