sin^2 theta -2 cos theta+1/4 =0
find value(°) of theta
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theta=60°
sin60°=√3/2
cos60°= 1/2
(√3/2)^2-2(1/2)+1/4
(3/4)-1+(1/4)
=0
sin60°=√3/2
cos60°= 1/2
(√3/2)^2-2(1/2)+1/4
(3/4)-1+(1/4)
=0
anant6:
can we do assumption of (°)
tell me man
leninviki
apply theta value 0,30,45,60°
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