How do you find all solutions to the equation in the interval [0,2pi] given cos2x - sinx = 0?

1 Answer
Binayaka C.
Aug 1, 2016

In the interval [0,2pi] Solution: x=30^0,150^0,270^0

Explanation:

cos2x-sinx=0 or 1-2sin^2x-sinx=0 (cos2x=1-2sin^2x) or 2sin^2x+sinx-1=0 or (2sinx-1)(sinx+1)=0 :. either (2sinx=1 ; sinx=1/2) or sinx =-1 When sinx =-1; x= 270^0. When sinx=1/2; x=30^0 , 150^0[Ans]

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