Rozwiąż równania

[gerla]

\(\displaystyle{ a) (sinx + cosx) ^{2} = cos2x}\)

\(\displaystyle{ b) sin ^{4} \frac{x}{3} + cos ^{4} \frac{x}{3} = \frac{5}{8}}\)

\(\displaystyle{ c) cos ^{4} x - sin ^{4} = sin4x}\)


w przykładzie c doszłam do tego:
\(\displaystyle{ cos ^{4} x - sin ^{4} = sin4x}\)
\(\displaystyle{ (cos ^{2} x + sin ^{2} x)(cos ^{2} x - sin ^{2} x) = sin4x}\)
\(\displaystyle{ 1 * cos2x = sin4x}\)
i dalej nie wiem..

[tometomek91]

a)
\(\displaystyle{ (sinx + cosx) ^{2} =cos^{2}x-sin^{2}x\\
(sinx + cosx) ^{2} =(cosx-sinx)(cosx+sinx)\\
(sinx + cosx) ^{2}-(cosx-sinx)(cosx+sinx)=0\\
(cosx+sinx)[(cosx+sinx)-(cosx-sinx)]=0\\
sinx+cosx=0 \vee (cosx+sinx)-(cosx-sinx)=0\\
\sqrt{2}(\frac{\sqrt{2}}{2}sinx+\frac{\sqrt{2}}{2}cosx)=0 \vee 2sinx=0\\
cos(45-x)=0 \vee sinx=0\\
...}\)

b)
\(\displaystyle{ sin ^{4} \frac{x}{3} + cos ^{4} \frac{x}{3} = \frac{5}{8}\\
sin ^{4} \frac{x}{3} + (cos ^{2} \frac{x}{3})^{2} = \frac{5}{8}\\
sin ^{4} \frac{x}{3}+(1-sin^{2} \frac{x}{3})^{2} = \frac{5}{8}\\
sin^{2} \frac{x}{3}=t\\
t^{2}+(1-t)^{2}=\frac{5}{8}\\
2t^{2}-2t+\frac{3}{8}=0\\
...}\)

c)
\(\displaystyle{ cos2x = sin4x\\
cos2x=2sin2xcos2x\\
cos2x-2sin2xcos2x=0\\
cosx(1-2sin2x)=0\\
cosx=0 \vee sin2x=\frac{1}{2}\\
...}\)