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sin theta= tan theta
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solve for theta?
solve the trig equation 0<theta<2pi
I replace theta =x for easy following sin x - tan x =0 \[sinx -\frac{sinx}{cosx}=0\] \[\frac{sinxcosx-sinx}{cosx}=0\] \[\frac{sinx (cosx -1)}{cosx}=0\] \[sinx =0 \]\[cosx -1=0\] \[sinx =0-----> x = 0 ; x = pi\] \[cosx -1=0 ----> cosx =1 ---> x = 0 ; x = 2pi\] therefore: x=0, x = pi and x =2pi are solutions
nope, no mistake
sin2xsinx=cosx
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what is it? the new one?
yes
sin2x = 2sinx cos x therefore 2sinxcosxsinx -cosx =0 cosx (2sin^2 x -1)=0 cos x =0 ---> x = pi/2 or 3pi/2 2sin^2x -1 = 0 ---> 2sin^2 x=1 ---> sin^2x = 1/2 ---> sin x = sqrt (1/2) x = pi/4 or 3pi/4
no more, friend
thank you so much!!
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