Rozwiąż równania (trygonometria)

[avleyi]

Rozwiąż:
a) \(sin(2x+ \frac{ \pi }{3}) = \frac{ \sqrt{3}}{2} \)

b) \(sin^2x+5sinx+4=0\)

c) \(sinx+ \sqrt{3}cosx= \sqrt{2} \)

d) \(cos3x+sin3x=-1\)

e) \( \frac{-x^2+x-2}{4sinxcosx-1} \le 0 \)

[kerajs]

a) \(sin(2x+ \frac{ \pi }{3}) = \frac{ \sqrt{3}}{2} \\
2x+ \frac{ \pi }{3}= \frac{ \pi }{3}+k 2\pi \ \ \ \ \vee \ \ \ \ 2x+ \frac{ \pi }{3}= \pi - \frac{ \pi }{3}+k 2\pi
\)

b) \(sin^2x+5sinx+4=0 \\
(\sin x+1)(\sin x +4)=0\\
\sin x=-1\\
x= \frac{- \pi }{2}+k2 \pi \)

c) \(sinx+ \sqrt{3}cosx= \sqrt{2} \\
2\sin (x+\frac{ \pi }{3})= \sqrt{2} \\
x+\frac{ \pi }{3}= \frac{ \pi }{4}+k 2\pi \ \ \ \ \vee \ \ \ \ x+\frac{ \pi }{3}= \pi - \frac{ \pi }{4}+k 2\pi \)

d) \(cos3x+sin3x=-1\\
\sqrt{2} \sin (3x+\frac{ \pi }{4})=-1\\
3x+\frac{ \pi }{4}=\frac{ - \pi }{4}+k 2\pi \ \ \ \ \vee \ \ \ \ 3x+\frac{ \pi }{4}= \pi - \frac{ - \pi }{4}+k 2\pi \)

e) \( \frac{-x^2+x-2}{4sinxcosx-1} \le 0 \\
\frac{(x+0,5)^2+1,75}{1-2\sin 2x} \le 0 \\
1-2\sin 2x<0\\
\sin 2x>0,5\\
\frac{ \pi }{6} +k2 \pi <2x < \pi -\frac{ \pi }{6} +k2 \pi
\)