Question

f(x)=2+3*cos((π/10)(x−3)) Find the first positive value of x for which f(x) = 4 huh?

Answer 1

$$
f(x)=2+3*cos\pmatrix{\cfrac{\pi}{10}(x−3)} \large \ ?
$$

Answer 2

i got 4.85

Answer 3

hmm, well how did you get that?

Answer 4

i put 4 in for x

Answer 5

"for which f(x) = 4"
^
"y"

Answer 6

$$
f(x)=2+3*cos\pmatrix{\cfrac{\pi}{10}(x−3)}\\
4=2+3*cos\pmatrix{\cfrac{\pi}{10}(x−3)}\\
\cfrac{2}{3} = cos\pmatrix{\cfrac{\pi}{10}(x−3)}\\
cos^{-1}\pmatrix{\cfrac{2}{3}} = cos^{-1}\pmatrix{cos\pmatrix{\cfrac{\pi}{10}(x−3)}}\\
\cfrac{2}{3} = \cfrac{\pi}{10}(x−3)
$$

Answer 7

well, I got a bit off hehe

Answer 8

$$
f(x)=2+3*cos\pmatrix{\cfrac{\pi}{10}(x−3)}\\
4=2+3*cos\pmatrix{\cfrac{\pi}{10}(x−3)}\\
\cfrac{2}{3} = cos\pmatrix{\cfrac{\pi}{10}(x−3)}\\
cos^{-1}\pmatrix{\cfrac{2}{3}} = cos^{-1}\pmatrix{cos\pmatrix{\cfrac{\pi}{10}(x−3)}}\\
48.19 = \cfrac{\pi}{10}(x−3)
$$

Answer 9

then you'd just need to solve for "x"

Answer 10

you're trying to solve
\[4 = 2+3\cos((\frac{\pi}{10})(x-3))\]Subtract 2 from both sides to give\[2=3\cos((\frac{\pi}{10})(x-3))\]Divide both sides by 3 to give\[\frac{2}{3}=\cos((\frac{\pi}{10})(x-3))\]Find the first value of \(\theta\) where \(\cos(\theta) = 2/3\) Then solve \(\theta = \pi/10(x-3)\) for \(x\)

Answer 11

i got .3141?

Answer 12

I got a much bigger degree number

Answer 13

well, degrees = 57.3 * radians

Answer 14

I get a number close to .3141, but I'm suspicious that you've just got pi/10 there :-)

Answer 15

ahemm, 0.3 radians is barely 18 degrees, I got a much bigger value

Answer 16

$$
\text{rounding off 48.19 to 48}\\
48 = \cfrac{\pi}{10}(x−3)\\
\cfrac{48\times 10}{\pi} = x-3
$$

Answer 17

you have to do the problem in radians...

Answer 18

hmm, I see, "4" as in 4 radians, :), well, just the 48 will change

Answer 19

so do youchange 48 to radians then solve?

Answer 20

This approach will get you a solution but it won't be the first positive value of x that makes it true.

Answer 21

Here's a plot of y=4 vs y=2+3cos[(pi/10)*(x-3)]