Question

How can I find the zeros for -3cos(3(x-(pi/4)))+2?

Answer 1

\[-3\cos\left(3\left(x-\frac{\pi}{4}\right)\right)+2=0
\\3\cos\left(3x-\frac{3\pi}{4}\right)=2
\\now\;solve\;for\;x\]

Answer 2

How?

Answer 3

@adrimit

Answer 4

Can you start by explaining what a zero is? Then we can help you figure out how to find the zeros of this particular expression.

Answer 5

It is where the graph crosses the x-axis

Answer 6

use the inverse function of cosine after isolating cos(...)=...

Answer 7

What?

Answer 8

@umerlodhi

Answer 9

Good, so let's get this in a form where you have cosine(something) on one side, and the rest of the number on the other. That is straight algebra.
Go ahead and set that up, @ohohaye please.

Answer 10

\[\cos \left( 3x-\frac{ 3\pi }{ 4 } \right)=\frac{ 2 }{ 3 }\]
can you expand it by the formula cos(a-b)=cos a cos b+sin a sin b?

Answer 11

@ohohaye , are you still working this problem? Your input here would be good.
@sshayer , yes, you have the equation correct, but let's let @ohohaye build up and show his work.
All y'all, although the formula that @sshayer gives is one we _could_ use, I think there is a quicker method. the cosine() function has an inverse that reverses it, called 'inverse cosine' or in older books 'arccosine'.
@ohohaye , what are you using to make calculations? A physical calculator or something on a phone or computer.

Answer 12

the easiest way is to use the following identity:
\[cos^{-1}(cos(f))=f
\\here,\;f\;is\;any\;function\;of\;x
\\once\;the\;equation\;is\;solved\;for\;cos\;f,\;multiply\;both\;sides\;by\;cos^{-1}
\\this\;is\;equivalent\;to\;taking\;the\;inverse\;of\;a\;function\;
\\so,\;if\;the\;equation\;is\;cos\left(3x-\frac{3\pi}{4}\right)=\frac{2}{3}
\\then\;the\;next\;step\;is\;cos^{-1}cos\left(3x-\frac{3\pi}{4}\right)=cos^{-1}\frac{2}{3}
\\next,3x-\frac{3\pi}{4}=cos^{-1}\frac{2}{3}
\\now\;solve\;for\;x\]

Answer 13

\[\cos \left( 3x-\frac{ 3 \pi }{ 4 } \right)=\frac{ 2 }{ 3 }\]
\[\cos 3x \cos \frac{ 3 \pi }{ 4 }+\sin 3x \sin \frac{ 3 \pi }{ 4 }=\frac{ 2 }{ 3 }\]
\[\cos 3x \cos \left( \pi-\frac{ \pi }{ 4 } \right)+\sin 3x \sin \left( \pi-\frac{ \pi }{ 4 } \right)=\frac{ 2 }{ 3 }\]
\[-\cos 3x \cos \frac{ \pi }{ 4 }+\sin 3x \sin \frac{ \pi }{ 4 }=\frac{ 2 }{ 3 }\]
\[-\frac{ 1 }{ \sqrt{2} }\cos 3x+\frac{ 1 }{ \sqrt{2} }\sin 3x=\frac{ 2 }{ 3 }\]
\[-\cos 3x+\sin 3x=\frac{ 2\sqrt{2} }{ 3 }\]
squaring both sides
\[\cos ^23x+\sin ^23x-2\sin 3x \cos 3x=\frac{ 8 }{ 9 }\]
\[1-\sin 6x=\frac{ 8 }{ 9 }\]
\[\sin 6x=1-\frac{ 8 }{ 9 }=\frac{ 1 }{ 9 }\]
\[6x=\sin^{-1} \left( \frac{ 1 }{ 9 } \right)=6.4 degree,180-6.4degree=6.4+360n,173.6+360 n\]
\[x=1.06+60n,28.9+60n\]
where n is an integer.
put n=0,1,2,3,4,5
and get the values.

Answer 14

wtf?