Question

Sin^2x+cosx-1=o
1-3cos^2x=2sinx*cos x

Answer 1

1.
$sin^{2} x+cosx-1=1- cos^{2} x+cosx-1=cosx-cos^{2}x = textgreater \ cosx - cos^{2}x=0 \ -cosx +cos^{2}x=0 \ cosx(cosx-1)=0 \ left[begin{array}{ccc}cosx=1\cosx=0end$
$x= pi n- pi /2 \ x=2 pi n$
2.
$1-3cos^{2}x=2sinxcosx|1=cos^2x+sin^2x \ sin^{2}x-2cos^{2}x-2sinxcosx=0 | :cos^{2}x \ tg^2x-2-2tgx=0 , ]tgx=n \ n^2-2n-2=0, D=4+8=12 \ n_{1} =1+ sqrt{3} \ n_{2}=1-sqrt3 \ tgx=1 frac{+}{} sqrt3 \ x=artctg(1 frac{+}{} sqrt3)+ pi k$
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