Nie potrafię sobie poradzić z tymi równaniami:
a) \(\displaystyle{ \sqrt{3} * cosx + sinx = 1}\)
\(\displaystyle{ tg \frac{ \pi}{3} * cosx + sinx = 1}\)
\(\displaystyle{ \frac{sin \frac{ \pi}{3}}{cos \frac{ \pi}{3}} * cosx + sinx = 1}\)
\(\displaystyle{ \frac{sin \frac{ \pi}{3}}{sin (\frac{ \pi}{3} + \frac{ \pi}{2} )} * sin (x+ \frac{ \pi}{2}) + sinx = 1}\) ...???
b) \(\displaystyle{ \frac{cosx}{1-sinx} = 1+sinx}\)
\(\displaystyle{ (1+sinx) (1-sinx) = cosx}\)
\(\displaystyle{ cos ^{2} x = cos}\)
\(\displaystyle{ \frac{cos ^{2}x }{cos} = 0}\)
\(\displaystyle{ cosx = 0}\)
\(\displaystyle{ x = \frac{\pi}{2}+ k \pi
x \epsilon \{ \frac{\pi}{2} +k \pi \}}\)
e) \(\displaystyle{ cos^{2} 7x = \frac{3}{4}}\)
\(\displaystyle{ cos^{2} 7x - \frac{3}{4} = 0}\)
\(\displaystyle{ (cos7x-\frac{3}{4})(cos7x+\frac{3}{4}) = 0}\) ...???
h) \(\displaystyle{ 4 sin ^{2} x + sin ^{2} 2x = 2}\) ...???
Byłabym wdzięczna za wyjaśnienie, co robię źle i jak to poprawnie rozwiązać?
Rozwiązać równania trygonometryczne
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JankoS
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Rozwiązać równania trygonometryczne
a) \(\displaystyle{ \sqrt{3} * cosx + sinx = 1}\)
\(\displaystyle{ tg \frac{ \pi}{3} * cosx + sinx = 1}\)
\(\displaystyle{ \frac{sin \frac{ \pi}{3}}{cos \frac{ \pi}{3}} * cosx + sinx = 1}\)
\(\displaystyle{ sin \frac{ \pi}{3}} \cdot cosx+{cos \frac{ \pi}{3}} \cdot sinx={cos \frac{ \pi}{3}}\\sin \left(\frac{\pi}{3}+x \right)=sin \left(\frac{\pi}{2}-x \right)\\\frac{\pi}{3}+x=\frac{\pi}{2}-x+2k\pi \ lub \ \frac {\pi}{3}+x=\pi-\frac{\pi}{2}-x+2k\pi.}\)
b) \(\displaystyle{ \frac{cosx}{1-sinx} = 1+sinx}\)
\(\displaystyle{ (1+sinx) (1-sinx) = cosx}\)
\(\displaystyle{ cos ^{2} x = cos}\)
\(\displaystyle{ cos ^{2} x - cosx=cosx(cosx-1)=0\\ cosx=0=cos\frac{\pi}{2} \ lub \ cosx=1=cos0\\\\x=\frac{\pi}{2}+k\pi \ lub \ x=2k\pi.}\)
e) \(\displaystyle{ cos^{2} 7x = \frac{3}{4}}\)
\(\displaystyle{ cos^{2} 7x - \frac{3}{4} = 0}\)
\(\displaystyle{ (cos7x-\frac{ \sqrt{3} }{2})(cos7x+\frac{ \sqrt{3} }{2}) =(cos7x-cos\frac{\pi}{6})(cos7x+cos\frac{\pi}{6}) = 0\\cos7x=cos\frac{\pi}{6} \ lub \ cos7x=-cos\frac{\pi}{6}=cos(\pi-\frac{\pi}{6})\\...}\)
h) \(\displaystyle{ 4 sin ^{2} x + sin ^{2} 2x = 2}\)
\(\displaystyle{ 4 sin ^{2} x +(2sinxcosx)^2=4 sin ^{2} x +4sin^2xcos^2x=2 \\ 2 sin ^{2} x +2sin^2xcos^2x-1=2 sin ^{2} x+2sin^2x(1-sin^2x)-1=\\=-2sin^4x+4sin^2x-1= 0\\2sin^4x-4sin^2x+1= 0\\ 0 \le t=sin^2x \le 1\\2t^2-4t+1=0}\)
itd.
\(\displaystyle{ tg \frac{ \pi}{3} * cosx + sinx = 1}\)
\(\displaystyle{ \frac{sin \frac{ \pi}{3}}{cos \frac{ \pi}{3}} * cosx + sinx = 1}\)
\(\displaystyle{ sin \frac{ \pi}{3}} \cdot cosx+{cos \frac{ \pi}{3}} \cdot sinx={cos \frac{ \pi}{3}}\\sin \left(\frac{\pi}{3}+x \right)=sin \left(\frac{\pi}{2}-x \right)\\\frac{\pi}{3}+x=\frac{\pi}{2}-x+2k\pi \ lub \ \frac {\pi}{3}+x=\pi-\frac{\pi}{2}-x+2k\pi.}\)
b) \(\displaystyle{ \frac{cosx}{1-sinx} = 1+sinx}\)
\(\displaystyle{ (1+sinx) (1-sinx) = cosx}\)
\(\displaystyle{ cos ^{2} x = cos}\)
\(\displaystyle{ cos ^{2} x - cosx=cosx(cosx-1)=0\\ cosx=0=cos\frac{\pi}{2} \ lub \ cosx=1=cos0\\\\x=\frac{\pi}{2}+k\pi \ lub \ x=2k\pi.}\)
e) \(\displaystyle{ cos^{2} 7x = \frac{3}{4}}\)
\(\displaystyle{ cos^{2} 7x - \frac{3}{4} = 0}\)
\(\displaystyle{ (cos7x-\frac{ \sqrt{3} }{2})(cos7x+\frac{ \sqrt{3} }{2}) =(cos7x-cos\frac{\pi}{6})(cos7x+cos\frac{\pi}{6}) = 0\\cos7x=cos\frac{\pi}{6} \ lub \ cos7x=-cos\frac{\pi}{6}=cos(\pi-\frac{\pi}{6})\\...}\)
h) \(\displaystyle{ 4 sin ^{2} x + sin ^{2} 2x = 2}\)
\(\displaystyle{ 4 sin ^{2} x +(2sinxcosx)^2=4 sin ^{2} x +4sin^2xcos^2x=2 \\ 2 sin ^{2} x +2sin^2xcos^2x-1=2 sin ^{2} x+2sin^2x(1-sin^2x)-1=\\=-2sin^4x+4sin^2x-1= 0\\2sin^4x-4sin^2x+1= 0\\ 0 \le t=sin^2x \le 1\\2t^2-4t+1=0}\)
itd.