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371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:06:03 | or \\
$h(t) = -4\sin \qty(\frac{\pi}{4(t-1)}) + 25$? | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 510,789,298,321,096,700 | TeXit | 01/30/2022 11:06:05 | **Ansh** | 706934233012371577.png | 0 | 1 | 510,789,298,321,096,700 | TeXit | 0796 | TeXit | true | ||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:06:10 | which one | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 692,553,195,230,068,700 | Looking123 | 01/30/2022 11:06:22 | this one | 1 | 0 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:06:30 | $h(t) = -4\sin \qty(\frac{\pi}{4}(t-1)) + 25$? | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 510,789,298,321,096,700 | TeXit | 01/30/2022 11:06:32 | **Ansh** | 706934233012371577.png | 0 | 1 | null | null | null | null | null | ||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 692,553,195,230,068,700 | Looking123 | 01/30/2022 11:06:37 | 1.859 | 1 | 0 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:07:01 | Okay, this is the height of the water wheel whatever **above the water surface** as a function of time "t" | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 692,553,195,230,068,700 | Looking123 | 01/30/2022 11:07:23 | below the water surface | 1 | 0 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 692,553,195,230,068,700 | Looking123 | 01/30/2022 11:07:50 | it is below from 1.86 to 4.14, | 1 | 0 | 510,789,298,321,096,700 | TeXit | 0796 | TeXit | true | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 692,553,195,230,068,700 | Looking123 | 01/30/2022 11:08:08 | I know how to get 1.85, I do not know how to get 4.14 | 1 | 0 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:08:09 | unknown.png | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 692,553,195,230,068,700 | Looking123 | 01/30/2022 11:08:36 | ohh that is cause I'm blind.....lol | 1 | 0 | 722,342,250,083,909,600 | ItzKar4n | 8310 | ItzKar4n | false | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:09:04 | Now, your question says.. that at a certain point of time, in the first 24 seconds( since t = 0 ), say at time t = k seconds, your h(k) = 0 | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 692,553,195,230,068,700 | Looking123 | 01/30/2022 11:09:12 | here I'll post an image, of what I don't understand] | 1 | 0 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:09:16 | SO basically, you solve for h(k) = 0 | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 692,553,195,230,068,700 | Looking123 | 01/30/2022 11:10:15 | unknown.png | 1 | 0 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:10:25 | Yeah I thought so | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 692,553,195,230,068,700 | Looking123 | 01/30/2022 11:10:32 | How do I algebraically find the circled | 1 | 0 | 706,934,233,012,371,600 | Ansh_ | 3502 | Arya | false | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:10:32 | there's many more out there lol | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:11:02 | So, $h(t) = -4\sin \qty(\frac{\pi}{4}(t-1)) + 2.5$ | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:12:14 | but this just pure blasphemy smh | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 510,789,298,321,096,700 | TeXit | 01/30/2022 11:12:32 | **ItzKar4n** | 722342250083909663.png | 0 | 1 | null | null | null | null | null | ||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:12:38 | okay wait 🤦♂️ | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 692,553,195,230,068,700 | Looking123 | 01/30/2022 11:12:44 | it is 2.5 | 1 | 0 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 510,789,298,321,096,700 | TeXit | 01/30/2022 11:12:50 | **Ansh** | 706934233012371577.png | 0 | 1 | null | null | null | null | null | ||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 510,789,298,321,096,700 | TeXit | 01/30/2022 11:12:55 | **Ansh**
```latex
$h(k) = 0 \implies -4\sin \qty(\frac{\pi}{4}(k-1)) + 2.5=0$
``` | 706934233012371577.png | 0 | 1 | null | null | null | null | null | ||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 692,553,195,230,068,700 | Looking123 | 01/30/2022 11:12:55 | yes | 1 | 0 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:13:31 | $4\sin \qty(\frac{\pi}{4}(k-1)) = 2.5$ | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 510,789,298,321,096,700 | TeXit | 01/30/2022 11:13:33 | **Ansh** | 706934233012371577.png | 0 | 1 | null | null | null | null | null | ||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:14:00 | $\implies \sin \qty(\frac{\pi}{4}(k-1)) = \frac{2.5}{4}$ | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 510,789,298,321,096,700 | TeXit | 01/30/2022 11:14:02 | **Ansh** | 706934233012371577.png | 0 | 1 | null | null | null | null | null | ||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:14:13 | Now comes the part that you're curious about | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:14:17 | :O | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:14:41 | We're given $k \in [0, 24]$ | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 510,789,298,321,096,700 | TeXit | 01/30/2022 11:14:43 | **Ansh** | 706934233012371577.png | 0 | 1 | null | null | null | null | null | ||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:15:31 | then, $\frac{\pi}{4}(k-1) \in \qty[-\frac{-\pi}{4}, \frac{23\pi}{4}]$ | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 510,789,298,321,096,700 | TeXit | 01/30/2022 11:15:43 | **Ansh** | 706934233012371577.png | 0 | 1 | null | null | null | null | null | ||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:16:05 | @Looking123 Any doubts so far? | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 692,553,195,230,068,700 | Looking123 | 01/30/2022 11:16:17 | Nope, I'm following | 1 | 0 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:16:29 | Perfect | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:16:34 | Note | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:16:43 | we had the equation: | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:16:49 | $\implies \sin \qty(\frac{\pi}{4}(k-1)) = \frac{2.5}{4}$ | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 510,789,298,321,096,700 | TeXit | 01/30/2022 11:16:52 | **Ansh** | 706934233012371577.png | 0 | 1 | 692,553,195,230,068,700 | Looking123 | 5951 | Looking123 | false | ||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:17:23 | which is basically $\sin \theta = \frac{2.5}{4}$ where $\theta \in \qty[-\frac{-\pi}{4}, \frac{23\pi}{4}]$ | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 510,789,298,321,096,700 | TeXit | 01/30/2022 11:17:26 | **Ansh** | 706934233012371577.png | 0 | 1 | null | null | null | null | null | ||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:18:32 | Notice, sine actually has a period of 2pi, so any solution you get for arcsin of something, is only a solution you got in the principal range of arcsin which is [ -pi/2 , pi/2 ] | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:19:09 | But here, you're interested in all solutions of theta for which this equation holds true, and not only those particular theta values between [-pi/2, pi/2] | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:19:36 | So, for starters | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:19:53 | You have: $\sin \theta = \frac{2.5}{4}$ | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 510,789,298,321,096,700 | TeXit | 01/30/2022 11:19:56 | **Ansh** | 706934233012371577.png | 0 | 1 | null | null | null | null | null | ||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:20:34 | First, consider a $\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ which is a solution of the given equation | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 510,789,298,321,096,700 | TeXit | 01/30/2022 11:20:37 | **Ansh** | 706934233012371577.png | 0 | 1 | null | null | null | null | null | ||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:21:16 | if theta lies in that interval, then you're safe to write down the next step: $\theta = \sin^{-1} \qty(\frac{2.5}{4})$ | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:21:51 | find this theta | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 510,789,298,321,096,700 | TeXit | 01/30/2022 11:22:05 | **Ansh** | 706934233012371577.png | 0 | 1 | null | null | null | null | null | ||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:22:09 | ,w arcsin(2.5/4) | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 510,789,298,321,096,700 | TeXit | 01/30/2022 11:22:13 | Results provided by WolframAlpha | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:22:22 | :this: | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:22:26 | You have this | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:23:10 | So, now you have a solution $\theta \in \qty[-\frac{\pi}{2}, \frac{\pi}{2}]$ to the above equation | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 510,789,298,321,096,700 | TeXit | 01/30/2022 11:23:26 | **Ansh** | 706934233012371577.png | 0 | 1 | null | null | null | null | null | ||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:23:39 | But who said this is the only solution to your original equation? | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 692,553,195,230,068,700 | Looking123 | 01/30/2022 11:23:48 | I don't think 0.675132 is a solution | 1 | 0 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:23:59 | uh, | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:24:06 | we are not finding "t" here | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:24:26 | we're just finding a solution to arcsin theta = 2.5/4 | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:24:38 | Yes you're misinterpreting lol | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:25:00 | Our "theta" is the term: $\qty(\frac{\pi}{4}(k-1))$ | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 692,553,195,230,068,700 | Looking123 | 01/30/2022 11:25:00 | ohh ok, lol. So do I need to add the period to 0.67 | 1 | 0 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 510,789,298,321,096,700 | TeXit | 01/30/2022 11:25:03 | **Ansh** | 706934233012371577.png | 0 | 1 | null | null | null | null | null | ||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:25:13 | Not only the period tbh | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:25:18 | Instead | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:25:31 | you notice that, for any equation $\sin x = c$ | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 510,789,298,321,096,700 | TeXit | 01/30/2022 11:25:34 | **Ansh** | 706934233012371577.png | 0 | 1 | null | null | null | null | null | ||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:25:44 | $\sin (\pi - x) = c$ also holds | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 510,789,298,321,096,700 | TeXit | 01/30/2022 11:25:46 | **Ansh** | 706934233012371577.png | 0 | 1 | null | null | null | null | null | ||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:26:00 | because $\sin (\pi - x) = \sin x = c$ | 0 | 1 | 692,553,195,230,068,700 | Looking123 | 5951 | Looking123 | false | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 510,789,298,321,096,700 | TeXit | 01/30/2022 11:26:03 | **Ansh** | 706934233012371577.png | 0 | 1 | null | null | null | null | null | ||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:26:10 | Hmm? | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:26:16 | Got it? | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 692,553,195,230,068,700 | Looking123 | 01/30/2022 11:26:27 | yeah, I think | 1 | 0 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:26:31 | yeah so | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:26:37 | the fault in your solution was | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:26:45 | you didn't account for these solutions of theta | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:27:03 | $\sin \theta = \frac{2.5}{4}$ where $\theta \in \qty[-\frac{-\pi}{4}, \frac{23\pi}{4}]$ | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 510,789,298,321,096,700 | TeXit | 01/30/2022 11:27:05 | **Ansh** | 706934233012371577.png | 0 | 1 | null | null | null | null | null | ||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 692,553,195,230,068,700 | Looking123 | 01/30/2022 11:27:16 | So using theta how would I get the solution now? | 1 | 0 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:27:30 | For $\theta \in \qty[-\frac{\pi}{2}, \frac{\pi}{2}], \theta = 0.675$ | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 510,789,298,321,096,700 | TeXit | 01/30/2022 11:27:33 | **Ansh** | 706934233012371577.png | 0 | 1 | null | null | null | null | null | ||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:28:48 | And for other intervals similarly, you use: $\theta, \pi - \theta, 2\pi + \theta, 3\pi - \theta, \dots$ until you reach the end of your interval of $\theta$ defined in the equation | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 510,789,298,321,096,700 | TeXit | 01/30/2022 11:29:05 | **Ansh** | 706934233012371577.png | 0 | 1 | null | null | null | null | null | ||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:29:39 | Now, you set aside these values and assume these solutions are just $\theta$ | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 510,789,298,321,096,700 | TeXit | 01/30/2022 11:29:45 | **Ansh** | 706934233012371577.png | 0 | 1 | null | null | null | null | null | ||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:30:01 | Now, you can go ahead and solve for "t" in terms of this \theta | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:30:21 | $\theta = \qty(\frac{\pi}{4}(k-1))$ | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 510,789,298,321,096,700 | TeXit | 01/30/2022 11:30:23 | **Ansh** | 706934233012371577.png | 0 | 1 | null | null | null | null | null | ||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 706,934,233,012,371,600 | Ansh_ | 01/30/2022 11:30:50 | $\implies k = 1+ \frac{4\theta}{\pi}$ | 0 | 1 | null | null | null | null | null | |||||
371 | help-10 | 3835d83e3a9d44b58bd0cba0d7d5babe | 510,789,298,321,096,700 | TeXit | 01/30/2022 11:30:52 | **Ansh** | 706934233012371577.png | 0 | 1 | null | null | null | null | null |
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