Question

cos(2x) less than or equal to 0 solve the inequality

Answer 1

\[\cos(2x) \le 0\]
first lets think about
\[\cos(x) \le 0\]
but even then lets first think about
\[\cos(x)=0\]
\[\cos(x)=0=> x=n \pi, n \in \mathbb{Z}\]

so this means \[\cos(2x)=0 when 2x=n \pi => x=\frac{n \pi}{2}\]

now let's think about \[\cos(x) <0\] on the interval \[(0,2\pi)\]
cos(x) is negative on the interval \[(\frac{\pi}{2}, \frac{3\pi}{2})\]
so
cos(2x) is negative on the interval \[(\frac{1}{2} \cdot \frac{\pi}{2}, \frac{1}{2} \cdot \frac{3\pi}{2})=(\frac{\pi}{4}, \frac{3\pi}{4})\]

but now we got to think for all real numbers that was just on the interval 0 to 2pi

so if we go around circle again
and we look at the interval 2pi to 4pi
we have that
cos(x) is negative on the interval \[(\frac{\pi}{2}+2\pi,\frac{3\pi}{2}+2\pi)\]
so cos(2x) would be negative on the interval
\[(\frac{1}{2}(\frac{\pi}{2}+2\pi),\frac{1}{2}(\frac{3\pi}{2}+2\pi))\]
so in general cos(2x) is negative on the interval(s):
\[(\frac{1}{2}(\frac{\pi}{2}+2n \pi)),\frac{1}{2}(\frac{3\pi}{2}+2npi))\]
if i didn't make a mistake

Answer 2

so
\[\cos(2x) \le 0\] when \[x=\frac{npi}{2} or x \in (\frac{1}{2}(\frac{\pi}{2}+2n \pi),\frac{1}{2}(\frac{3\pi}{2}+2n \pi))\]

Answer 3

\[x=\frac{n \pi}{2} or x \in (\frac{1}{2}(\frac{\pi}{2}+2n \pi),\frac{1}{2}(\frac{3\pi}{2}+2n \pi))\]

Answer 4

Im confused so cos = \[\sqrt{2}\div2\] and i have to jus all reference angles?

Answer 5

what?

Answer 6

i get it thanks

Answer 7

could you help me with find the inverse of F(x)=x^3+9x7

Answer 8

is that 7th power?

Answer 9

\[x ^{3}+9x+7\]