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t+√(1-t²)=1
√(1-t²)=1-t |²
1-t²=1-2t+t²
-t²-t²-2t+1-1=0
-2t²-2t=0
-2t(t+1)=0
-2t=0 t=-1
t=0
sinx=0 lub sinx=-1
Odp: x należy do {2kpi suma -pi/2+2kpi; k należy do całkowitych}