Jak rozwiązać równanie
\(\displaystyle{ cos(x + \frac{pi}{3}) + sin( \frac{pi}{6}- x) = \sqrt{2}}\)
najbardziej zalezy mi na informacji jak zmienic ten \(\displaystyle{ sin( \frac{pi}{6}- x)}\) na cos?
Równanie trygonometryczne łatwe
Równanie trygonometryczne łatwe
Ostatnio zmieniony 7 lut 2010, o 23:26 przez xanowron, łącznie zmieniany 2 razy.
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Równanie trygonometryczne łatwe
\(\displaystyle{ \cos\left(x + \frac{\pi}{3}\right) + \sin\left( \frac{\pi}{6}- x\right) = \sqrt{2}}\)
\(\displaystyle{ \pi=180^\circ}\)
\(\displaystyle{ \sin \left( \frac{\pi}{6}- x\right) =\sin \left( \frac{\pi}{6}\right) \cos \left(- x \right) +\cos \left( \frac{\pi}{6}\right) \sin \left( - x\right)\\
\sin \left( \frac{\pi}{6}- x\right) =\sin \left( \frac{180^\circ}{6}\right) \cos \left(- x \right) +\cos \left( \frac{180^\circ}{6}\right) \sin \left( - x\right)\\
\sin \left( \frac{\pi}{6}- x\right) = \sin \left( 30^\circ\right) \cos \left(- x \right) +\cos \left( 30^\circ\right) \sin \left( - x\right)\\
\sin \left( \frac{\pi}{6}- x\right) = \frac{1}{2} \cos \left(- x \right) + \frac{ \sqrt{3} }{2} \sin \left( - x\right)}\)
\(\displaystyle{ \cos \left( x+\frac{\pi}{3}\right) =\cos \left( x\right) \cos \left(\frac{\pi}{3} \right) -\sin \left(x\right) \sin \left( \frac{\pi}{3}\right)\\
\cos \left( x+\frac{\pi}{3}\right) =\cos \left( x\right) \cos \left(\frac{180^\circ}{3} \right) -\sin \left(x\right) \sin \left( \frac{180^\circ}{3}\right)\\
\cos \left( x+\frac{\pi}{3}\right) =\cos \left( x\right) \cos \left(60^\circ \right) -\sin \left(x\right) \sin \left( 60^\circ\right)\\
\cos \left( x+\frac{\pi}{3}\right) = \frac{1}{2} \cos \left( x\right) -\frac{ \sqrt{3} }{2}\sin \left(x\right)}\)
\(\displaystyle{ \cos \left(- x \right)= \cos \left(x \right)\\
\sin \left( - x\right)=-\sin \left(x\right)}\)
\(\displaystyle{ \cos\left(x + \frac{\pi}{3}\right) + \sin\left( \frac{\pi}{6}- x\right) = \sqrt{2}\\
\frac{1}{2} \cos \left( x\right) -\frac{ \sqrt{3} }{2}\sin \left(x\right)+\frac{1}{2} \cos \left(x \right) - \frac{ \sqrt{3} }{2} \sin \left(x\right)= \sqrt{2}\\
\cos \left( x\right) -\sqrt{3}\sin \left(x\right)= \sqrt{2}\\
\cos \left( x\right) = \sqrt{2}+\sqrt{3}\sin \left(x\right)\\
\cos^2 \left( x\right) = 2+2\sqrt{6}\sin \left(x\right)+3\sin^2 \left(x\right)\\
0 = 2+2\sqrt{6}\sin \left(x\right)+3\sin^2 \left(x\right)-\cos^2 \left( x\right)\\
0 = 1+2\sqrt{6}\sin \left(x\right)+3\sin^2 \left(x\right)+1-\cos^2 \left( x\right)\\
0 = 1+2\sqrt{6}\sin \left(x\right)+3\sin^2 \left(x\right)+sin^2 \left(x\right)\\
0 = 1+2\sqrt{6}\sin \left(x\right)+4\sin^2 \left(x\right)\\
\sin \left(x\right)=y\\
0 = 1+2\sqrt{6}y+4y^2\\
\Delta=24-16=8\\
y_1= \frac{-2\sqrt{6}-2\sqrt{2} }{8}=\frac{-\sqrt{6}- \sqrt{2} }{4} \approx -0,9659\\
y_2= \frac{-2\sqrt{6}+ 2\sqrt{2} }{8} =\frac{-\sqrt{6}+ \sqrt{2} }{4} \approx -0,2588\\
x_1=-75^\circ\\
x_2=-15^\circ}\)
\(\displaystyle{ \pi=180^\circ}\)
\(\displaystyle{ \sin \left( \frac{\pi}{6}- x\right) =\sin \left( \frac{\pi}{6}\right) \cos \left(- x \right) +\cos \left( \frac{\pi}{6}\right) \sin \left( - x\right)\\
\sin \left( \frac{\pi}{6}- x\right) =\sin \left( \frac{180^\circ}{6}\right) \cos \left(- x \right) +\cos \left( \frac{180^\circ}{6}\right) \sin \left( - x\right)\\
\sin \left( \frac{\pi}{6}- x\right) = \sin \left( 30^\circ\right) \cos \left(- x \right) +\cos \left( 30^\circ\right) \sin \left( - x\right)\\
\sin \left( \frac{\pi}{6}- x\right) = \frac{1}{2} \cos \left(- x \right) + \frac{ \sqrt{3} }{2} \sin \left( - x\right)}\)
\(\displaystyle{ \cos \left( x+\frac{\pi}{3}\right) =\cos \left( x\right) \cos \left(\frac{\pi}{3} \right) -\sin \left(x\right) \sin \left( \frac{\pi}{3}\right)\\
\cos \left( x+\frac{\pi}{3}\right) =\cos \left( x\right) \cos \left(\frac{180^\circ}{3} \right) -\sin \left(x\right) \sin \left( \frac{180^\circ}{3}\right)\\
\cos \left( x+\frac{\pi}{3}\right) =\cos \left( x\right) \cos \left(60^\circ \right) -\sin \left(x\right) \sin \left( 60^\circ\right)\\
\cos \left( x+\frac{\pi}{3}\right) = \frac{1}{2} \cos \left( x\right) -\frac{ \sqrt{3} }{2}\sin \left(x\right)}\)
\(\displaystyle{ \cos \left(- x \right)= \cos \left(x \right)\\
\sin \left( - x\right)=-\sin \left(x\right)}\)
\(\displaystyle{ \cos\left(x + \frac{\pi}{3}\right) + \sin\left( \frac{\pi}{6}- x\right) = \sqrt{2}\\
\frac{1}{2} \cos \left( x\right) -\frac{ \sqrt{3} }{2}\sin \left(x\right)+\frac{1}{2} \cos \left(x \right) - \frac{ \sqrt{3} }{2} \sin \left(x\right)= \sqrt{2}\\
\cos \left( x\right) -\sqrt{3}\sin \left(x\right)= \sqrt{2}\\
\cos \left( x\right) = \sqrt{2}+\sqrt{3}\sin \left(x\right)\\
\cos^2 \left( x\right) = 2+2\sqrt{6}\sin \left(x\right)+3\sin^2 \left(x\right)\\
0 = 2+2\sqrt{6}\sin \left(x\right)+3\sin^2 \left(x\right)-\cos^2 \left( x\right)\\
0 = 1+2\sqrt{6}\sin \left(x\right)+3\sin^2 \left(x\right)+1-\cos^2 \left( x\right)\\
0 = 1+2\sqrt{6}\sin \left(x\right)+3\sin^2 \left(x\right)+sin^2 \left(x\right)\\
0 = 1+2\sqrt{6}\sin \left(x\right)+4\sin^2 \left(x\right)\\
\sin \left(x\right)=y\\
0 = 1+2\sqrt{6}y+4y^2\\
\Delta=24-16=8\\
y_1= \frac{-2\sqrt{6}-2\sqrt{2} }{8}=\frac{-\sqrt{6}- \sqrt{2} }{4} \approx -0,9659\\
y_2= \frac{-2\sqrt{6}+ 2\sqrt{2} }{8} =\frac{-\sqrt{6}+ \sqrt{2} }{4} \approx -0,2588\\
x_1=-75^\circ\\
x_2=-15^\circ}\)