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Sa se rezolve in multimea numerelor reale ecuatia sinx + cos(-x)=1

Răspuns :

impartim expresia la radical din 2 si obtinem 1/radical din 2 sinx+1/radical in 2 cos x,dupa cum vezi expresia este egala cu 1/radical din 2
sinx+cosx=1
[tex]sin2 \frac{x}{2}+cos2 \frac{x}{2}=cos^2 \frac{x}{2}+sin^2 \frac{x}{2} \\ 2sin \frac{x}{2}cos \frac{x}{2}+cos^2 \frac{x}{2}-sin^2 \frac{x}{2}-cos^2 \frac{x}{2}-sin^2 \frac{x}{2}=0 \\ 2sin \frac{x}{2}cos \frac{x}{2}-2sin^2 \frac{x}{2}=0 \\ 2sin \frac{x}{2}(cos \frac{x}{2}-sin \frac{x}{2})=0 \\ sin \frac{x}{2}=0;sau;cos \frac{x}{2}-sin \frac{x}{2}=0 \\ \frac{x}{2}=k \pi \\ \frac{x}{2}=n \pi ;x=2n \pi [/tex]