trygonometria- równania, nierówności itp.

[uczen0313]

1.Rozwiąż równania:
a) \(\displaystyle{ sin2x=1}\)
b) \(\displaystyle{ sin(2x- \frac{ \pi }{6})= \frac{1}{2}}\)

2.
Rozwiąż nierówność:
\(\displaystyle{ tgx> \sqrt{3}}\)

3. (zad. wyboru)
Jeśli dla pewnego kąta ostrego \(\displaystyle{ \alpha}\) mamy \(\displaystyle{ cos \alpha =2M}\), to:
a) \(\displaystyle{ sin \alpha = \sqrt{1-4M ^{2} }}\)

b)\(\displaystyle{ tg \alpha = \frac{2M}{ \sqrt{1-4M ^{2} } }}\)

c) \(\displaystyle{ sin \alpha = \frac{1}{ \sqrt{1-4M ^{2} } }}\)

d) \(\displaystyle{ tg \alpha = \frac{ \sqrt{1-4M ^{2} } }{2M}}\)

4.
a)\(\displaystyle{ sinx= \frac{ \sqrt{2} }{2}}\)
x=? lub x=?
b) \(\displaystyle{ tg3x=1}\)
x=?

[bartek118]

1.
a)
\(\displaystyle{ 2x= \frac{\pi }{2} + 2k \pi}\)
\(\displaystyle{ x= \frac{\pi }{4} + k \pi}\)
b)
\(\displaystyle{ 2x- \frac{ \pi }{6} = \frac{ \pi }{6} + 2k \pi}\) lub \(\displaystyle{ 2x- \frac{ \pi }{6}=\frac{5 \pi }{6} + 2k\pi}\)
\(\displaystyle{ 2x= \frac{ \pi }{3} + 2k\pi}\) lub \(\displaystyle{ 2x=\pi + 2k\pi}\)
\(\displaystyle{ x=\frac{ \pi }{6}+k\pi}\) lub \(\displaystyle{ x= \frac{\pi}{2} +k\pi}\)
2.
\(\displaystyle{ \frac{1}{3} (3k \pi +\pi)<x< \frac{1}{2} (2k \pi +\pi)}\)
3.
\(\displaystyle{ cos^{2}\alpha=4M^{2}}\)
\(\displaystyle{ 1-sin^{2}\alpha=4M^{2}}\)
\(\displaystyle{ sin\alpha=\sqrt{1-4M^{2}}}\)
\(\displaystyle{ tg\alpha=\frac{\sqrt{1-4M^{2}}}{2M}}\)
4.
a)
\(\displaystyle{ x= \frac{\pi}{4} +2k\pi}\) lub \(\displaystyle{ x=\frac{3\pi}{4}+2k\pi}\)
b)
\(\displaystyle{ x= \frac{\pi}{12}+ \frac{k\pi}{3}}\)