Question

At which values in the interval [0, 2π) will the functions f (x) = cos 2x - 5 and g(x) = sin3x - 5 intersect?

A.
x = 0, π/10, π/2, 9π/10, 13π/10

B.
x = 3π/10, π/2, 7π/10, 13π/10, 17π/10

C.
x = π/10, π/2, 9π/10, 13π/10, 17π/10

D.
x = 0, π/2, π/10, 13π/10, 17π/10

Answer 1

I thinking option B.)

Answer 2

Yep that should be correct

Answer 3

You sure?

Answer 4

@surjithayer

Answer 5

let me check

Answer 6

yep seems right u can still check with someone else first

Answer 7

Okay thanks

Answer 8

\[\cos 2x-5=\sin 3x-5\]
\[\cos 2x=\sin 3x\]
\[\cos 2x=\cos (\frac{ \pi }{ 2 }-3x)\]
\[\cos 2x-\cos (\frac{ \pi }{ 2 }-3x)=0\]
\[\cos C-\cos D=-2\sin \frac{ C+D }{ 2 }\sin \frac{ C-D }{2 }\]
\[-2\sin \frac{ 2x+\frac{ \pi }{ 2 } -3x}{ 2}\sin \frac{ 2x-\frac{ \pi }{ 2 } +3x}{ 2} =0\]
either
\[\sin (\frac{ \pi }{ 4 }-\frac{ x }{ 2 })=0=\sin (n \pi)\]
\[\frac{ \pi }{ 4 }-\frac{ x }{ 2 }=n \pi \]
\[\pi-2x=4n \pi \]
\[2x=\pi-4n \pi \]
n=0
\[x=\frac{ \pi }{ 2 }\]
n=-1
\[x=\frac{ 5\pi }{ 2 },rejected\]
\[x \in[0,2\pi]\]

or
\[\sin (\frac{ 5x }{ 2 }-\frac{ \pi }{ 4 })=0=\sin n \pi \]
\[\frac{ 5x }{ 2 }-\frac{ \pi }{ 4 }=n \pi,n\in I\]
\[10x-\pi=4n \pi\]\[10x=4n \pi+\pi \]
n=0
\[x=\frac{ \pi }{ 10 }\]
n=1
\[x=\frac{ 5\pi }{ 10 }=\frac{ \pi }{ 2 }\]
n=2
\[10x=9\pi \]\[x=\frac{ 9\pi }{ 10}\]
n=3
\[10x=13\pi \]\[x=\frac{ 13 \pi }{ 10 }\]
n=4
\[10x=17\pi \]\[x=\frac{ 17\pi }{ 10}\]
n=5
\[10x=20\pi+\pi=21\pi\]\[x=\frac{ 21\pi }{ 10 }\]\[Rejected\]

Answer 9

so C

Answer 10

Ohhh I see wow thank you sm!