Question

solve sin² theta - cos theta = 1 by 4 for theta and write values of theta in interval 0< theta<2π

Answer 1

sin^2 theta - cos theta=1/4

=>1-cos^2 theta-cos theta=1/4

=>cos^2theta - cos theta -3/4=0
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by formula
x=[-b±√b^2-4ac]/2a


cos theta=[-1±√{1-+4(1)(3/4)}]/2

cos theta =[-1±√(1+3)]/2

cos theta=(-1±2)/2

cos theta =1/2,-3/2

but -1≤cos theta≤1

so cos theta=1/2
or
cos theta=cos(π/3) or cos 4π/3

as theta belongs to 0 to 2π

theta =π/3,4π/3

Answer 2

π/3

4π/3


these are the answers


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