sinx + cos x + sin2x = 1
sin x + cos x + 2sinx cosx -1=0
sin x + cos x +2sinx cosx -(sin²x+cos²x)=0
(sin x + cos x) + 2sinx cos x - (sin²x+cos²x+2sinx cosx -2sinx cos x)=0
(sin x+ cos x)+2sinx cosx - (sin x + cos x)² +2sinx cosx=0
(sin x + cos x)² + (sinx + cosx)+4sinxcosx=0
Пусть sin x + cos x = t причем (-√2 ≤ t ≤ √2), тогда возведем оба части до квадрата, имеем
(sin x + cos x)² = t²
1+2sinx cosx = t²
2sinxcosx = t²-1
Заменяем
t²+t+2*(t²-1)=0
t²+t+2t²-2=0
3t²+t-2=0
D=1+24 = 25
t1=(-1+5)/6=2/3
t2=(-1-5)/6 = -1
Возвращаем к замене
\begin{gathered}\sin x+\cos =-1\\ \sqrt{2} \sin(x+ \frac{\pi}{4} )=-1 \\ \sin(x+ \frac{\pi}{4} )=- \frac{1}{ \sqrt{2} } \\ x+ \frac{\pi}{4}=(-1)^{n+1} \frac{\pi}{4}+ \pi n,n \in Z\\ x=(-1)^{n+1} \frac{\pi}{4}- \frac{\pi}{4}+ \pi n,n \in Z\end{gathered}
sinx+cos=−1
2
sin(x+
4
π
)=−1
sin(x+
4
π
)=−
2
1
x+
4
π
=(−1)
n+1
4
π
+πn,n∈Z
x=(−1)
n+1
4
π
−
4
π
+πn,n∈Z
\begin{gathered}\sin x+\cos x= \frac{2}{3} \\ \sqrt{2} \sin(x+ \frac{\pi}{4})= \frac{2}{3} \\ \sin (x+ \frac{\pi}{4})= \frac{ \sqrt{2} }{3} \\ x=(-1)^n\arcsin( \frac{ \sqrt{2} }{3} )- \frac{\pi}{4}+ \pi n,n \in Z\end{gathered}
sinx+cosx=
3
2
2
sin(x+
4
π
)=
3
2
sin(x+
4
π
)=
3
2
x=(−1)
n
arcsin(
3
2
)−
4
π
+πn,n∈Z
