trigonometry

G

Guest

Select parameter a ;

\(\displaystyle \sin^{4}x-2\cos^{2}x+a^{2}=0\)
 
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Oct 2011
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Belgium
noki said:
Select parameter a ;

\(\displaystyle \sin^{4}x-2\cos^{2}x+a^{2}=0\)
\(\displaystyle (1-\cos^{2}x)^{2}-2\cos^{2}x+a^{2}=0\)
\(\displaystyle \cos^{4}x-4\cos^{2}x+1+a^{2}=0\)
\(\displaystyle \left(\frac{\cos2x+1}{2}\right)^2-4\frac{\cos2x+1}{2}+1+a^{2}=0\)
\(\displaystyle \cos^{2}(2x)-4\cos(2x)-3+4a^{2}=0\)
\(\displaystyle \cos(2x)=2+-\sqrt{7-4a^{2}}\)
\(\displaystyle 1<=\sqrt{7-4a^{2}}<=3\)
\(\displaystyle 1<=7-4a^{2}<=9\)
\(\displaystyle -6<=-4a^{2}<=3\)
\(\displaystyle -3/2<=a<=3/2\)
 
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G

Guest

thanks a lot
 

skipjack

Forum Staff
Dec 2006
21,394
2,413
Unfortunately, wnvl made a couple of slips.

The original equation is equivalent to $(\sin^2(x) + 1)^2 + (a^2 - 3) = 0$,
so (for $x$ to be real) $a^2 \leqslant 2$, i.e. $-√2 \leqslant a \leqslant √2$.