Question

5=1+5sin(pi/2)(x+0.7)

I'm supposed to find the first three positive values. I know how to do this with degrees, but I'm stuck when it comes to doing it in radians.

Answer 1

\[5=1+5 \cdot \sin(\frac{\pi}{2}) \cdot (x+\frac{7}{10}) ?\]

Answer 2

\[5=1+5\sin \frac{ \pi }{ 2 } (x+0.7)\]

Answer 3

what is that sin function operating on?

Answer 4

just the pi/2

Answer 5

or pi/2(x+7/10)?

Answer 6

\[5=1+5 \cdot \sin(\frac{\pi}{2}(x+\frac{7}{10}) ) ? \\ \text{ I think you mean this }\]

Answer 7

First isolate the part that is \[\sin(\frac{\pi}{2} \cdot (x+\frac{7}{10}) ) \]

Answer 8

No the one I typed out in my response is the one I have in my textbook. I've gotten as far as \[2\pi n \pm 0.93 = \frac{ \pi }{ 2 }(x+0.7)\]

Answer 9

I'm supposed to divide both sides by \[\frac{ \pi }{ 2 }\] but I keep getting the wrong answer after that.

Answer 10

try multiplying by the reciprocal of pi/2 on both sides instead of dividing both sides by pi/2
looks at like that might make it less complicated for you

Answer 11

Is that supposed to give me \[4\pi \pm 1.86\]?

Answer 12

\[\frac{2}{\pi}(2 \pi n \pm 0.93)=x+0.7\]

Answer 13

and no

Answer 14

\[4\pi \pm 0.592\]? because I got that and tried to complete the equation from there and I still got the wrong answer. Not really sure what I'm doing incorrectly.

Answer 15

well i just said no

Answer 16

\[\frac{2}{\pi}(2 \pi n \pm 0.93)=x+0.7 \\ \text{ distribute } \\ \frac{2 \cdot 2 \pi n }{ \pi } \pm \frac{2}{\pi} 0.93=x+0.7\]

Answer 17

the pi's would cancel in the first term

Answer 18

also do you know that sin(-0.93) isn't 4/5

Answer 19

I was supposed to take the arcsin of 4/5 which is 0.93

Answer 20

right sin(.93) is 4/5
but sin(-.93) isn't 4/5

Answer 21

anyways .93 is between 0 and pi/2
so another number that occurs between 0 and 2pi
pi-.93 is another that occurs between 0 and 2pi such that the sin value is 4/5
I know this from looking at the unit circle

Answer 22

So this means you would have sin(.93+2pi*n)=4/5
or sin((pi-.93)+2pi*n)=4/5

Answer 23

where n is an integer

Answer 24

But anyways you still need to solve
\[.93+2 \pi \cdot n =\frac{\pi}{2}(x+.7) \\ \text{ and } \pi-.93+2\pi \cdot n=\frac{\pi}{2}(x+.7)\]