Exercise:
Solve in $\mathbb{R}$,
1)
$\sqrt{x}-\left| x-2 \right|<1$
2)
${{\sin }^{4}}x+{{\cos }^{4}}x=1$
3)
$\sum\nolimits_{i=0}^{5}{{{\left( -1
\right)}^{i+1}}{{x}^{i}}=0}$
Solution:
1)
We know that, if $\sqrt{f\left( x \right)}<g\left(
x \right)$ then $g\left( x \right)>0$
So
that, $1+\left| x-2 \right|>0\Leftrightarrow \left| x-2 \right|>-1$ true for all $x$
Notice that, the domain of this equation $x\ge 0$
So
that there are 2 cases considered in this study
Case
1: $x>2$
$\sqrt{x}-\left|
x-2 \right|<1\Leftrightarrow \sqrt{x}-x+2<1\Leftrightarrow
\sqrt{x}-x+1<0\Leftrightarrow {{x}^{2}}-3x+1>0$
So
that, $x>\frac{3+\sqrt{5}}{2}$
Case
2: $x<2$
$\sqrt{x}-\left|
x-2 \right|<1\Leftrightarrow \sqrt{x}+x-2<1\Leftrightarrow
\sqrt{x}+x-3<0\Leftrightarrow {{x}^{2}}-7x+9>0$
So that,
$0\le x<\frac{7-\sqrt{13}}{2}$
2)
We have ${{\sin }^{4}}x+{{\cos
}^{4}}x=1\Leftrightarrow {{\left( {{\sin }^{2}}x \right)}^{2}}+{{\left( {{\cos
}^{2}}x \right)}^{2}}=1$
$\Leftrightarrow
{{\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)}^{2}}=1+2{{\sin }^{2}}x{{\cos
}^{2}}x\Leftrightarrow 2{{\sin }^{2}}x{{\cos }^{2}}x=0$
$\Leftrightarrow
\sqrt{2}\sin x\cos x=0\Leftrightarrow \sin x=0\,\,or\,\,\,\cos
x=0\Leftrightarrow x=0\,\,or\,\pi +2k\pi \,\,or\,\,x=\pm \frac{\pi }{2}+2k\pi
\,\,,\,\,k\in \mathbb{Z}$
Or
we can use factorization without going complete squaring as follows
${{\sin
}^{4}}x+{{\cos }^{4}}x=1\Leftrightarrow {{\cos }^{4}}x=1-{{\sin }^{4}}x=\left(
1-{{\sin }^{2}}x \right)\left( 1+{{\sin }^{2}}x \right)={{\cos }^{2}}x\left(
1+{{\sin }^{2}}x \right)$
so
that, ${{\cos }^{2}}x\left[ {{\cos }^{2}}x-\left( 1+{{\sin }^{2}}x \right)
\right]=0\Leftrightarrow {{\cos }^{2}}x\left( \cos 2x-1 \right)=0$
$\Leftrightarrow x=\pm \frac{\pi }{2}+2k\pi
\,\,or\,\,x=0+k\pi $ with $k\in \mathbb{Z}$
3)
$S=\sum\nolimits_{i=0}^{5}{{{\left( -1
\right)}^{i+1}}{{x}^{i}}=-1+x-{{x}^{2}}+{{x}^{3}}-{{x}^{4}}+{{x}^{5}}}=0$
Note
that, the above series is geometric with common ratio is $-x$ and initial value
$-1$
So that, $S=\sum\nolimits_{i=0}^{n-1}{a{{r}^{i}}}=-\left( \frac{1-{{\left( -x \right)}^{6}}}{1+x} \right)=0\Leftrightarrow 1-{{x}^{6}}=0\Leftrightarrow {{x}^{6}}=1\Leftrightarrow x=1$
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