Question

Trigonometry question

Answer 1

\(\color{#000000 }{ \displaystyle \frac{1}{x}+x\ge 1\quad \forall x }\)

Answer 2

Oh, my bad, I meant

\(\color{#000000 }{ \displaystyle \left|\frac{1}{x}+x\right|\ge 1\quad \forall x }\)

Answer 3

Yes yes...

Answer 4

So, there is your answer ;)

Answer 5

i think u meant

\(\large \color{black}{\begin{align}
& x+\dfrac1x\geq 2\hspace{.33em}\\~\\
\end{align}}\)

Answer 6

what to do next

Answer 7

You know that
\(\color{#000000 }{ \displaystyle \left|\sin x\right|\le 1 }\) and \(\color{#000000 }{ \displaystyle \left|\cos x\right|\le 1 }\)

(1st eq, not relevant here)

Answer 8

\(\color{#000000 }{ \displaystyle \cos \theta=\frac{1}{x}+x }\)

is an equivalent to saying,

\(\color{#000000 }{ \displaystyle \cos \theta=2{\rm ~or~more} }\)

Answer 9

well, for negatives , -2 or less.

Answer 10

the point is tho, that cosine (and sine) are always between -1 and 1.

Answer 11

is it

\(\large \color{black}{\begin{align}
& d.)\ \hspace{.33em}\\~\\
\end{align}}\)

Answer 12

yes

Answer 13

Yes, it is d.

\(d\theta\) :)

Answer 14

too late :)

Answer 15

Because parth replied, or because you submitted?

Answer 16

i m not on a test to submit :)

Answer 17

cu then, then.... :)

Answer 18

Some things are better to be kept unspoken -joker

Answer 19

\( ~~~\)

Answer 20

\[x+\frac{ 1 }{ x }=\cos \theta ,x^2-x \cos \theta +1=0\]
\[x=\frac{ \cos \theta \pm \sqrt{\left( -\cos \theta \right)^2-4*1*1} }{ 2*1 }=\frac{ \cos \theta \pm \sqrt{\cos ^2 \theta -4} }{ 2 }\]
x is real if \[\cos ^2\theta -4\geq 0 ~or \left| \cos \theta \right| \geq2\]
which is impossible as \[\left| \cos \theta \right|\leq 1\]
Hence x has no real solution.