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$$4\sin^2\left(\frac\pi5(x-1)\right)=3\qquad x\in[0 , 2π)$$

My work:

$$\sin^2[\pi/5(x-1)] = 3/4$$

Let $p = \pi/5(x-1)$

$$\sin^2 p = 3/4$$

$$\sin p = \pm \sqrt{3}/2$$

$$p = \pi/3, 2\pi/3, 4\pi/3, 5\pi/3$$

$$\pi/5(x-1) = \pi/3, 2\pi/3, 4\pi/3, 5\pi/3$$

$$x = 8/3, 13/3, 23/3, 28/3$$

However, I'm not sure if these answers are correct or how many may be correct answers within the restricted range.

If, hypothetically, I wanted to set the range of $p$, what would it be without finding the correct $x$ values first?

Can you show how you would do the problem?

andrew chen
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  • You have not found the values for x in the required range because you picked a limited set of values for p with no concern for the values of x. – William Elliot Jan 13 '19 at 03:44
  • @WilliamElliot Could you show me how to get the full range of values for p in order to get all the x values within the restricted range? – andrew chen Jan 13 '19 at 03:55
  • @andrewchen: $0\le x<2\pi$ implies $a\le\frac{\pi}{5}(x-1)<b$ for some $a$ and $b$. Can you find what they are? –  Jan 13 '19 at 04:02
  • @user587192 ohh so would it then be -π/5 ≤ p < π/5(2π-1)? In the final answer, would there then be only 2 values of x? – andrew chen Jan 13 '19 at 04:22
  • @andrewchen: your inequality is correct. Can you find out what values of $p$ would work? And then you have the corresponding values of $x$. –  Jan 13 '19 at 15:15
  • @user587192 out of the 4 possible values, there only values that could work would be pi/3 and 2pi/3. Thanks for your help! – andrew chen Jan 13 '19 at 16:57
  • @andrewchen: I'm not sure why you say "4 possible values". It seems that you restricted your $p$ to be in $[0,2\pi]$ originally. But one should consider all "possible" $p$ such that $|\sin p|=\frac{\sqrt{3}}{2}$, which means $p=\frac{\pi}{3}+k\pi$ or $p=\frac{{2}}{3}\pi+k\pi$ where $k$ is an integer. And then you search among these values that are in the interval $[-\pi/5,\pi(2\pi-1)/5)$. –  Jan 13 '19 at 17:31
  • @user587192 ok, that makes sense. – andrew chen Jan 13 '19 at 17:54

3 Answers3

1

Since $x\in[0,2\pi)$, you must have $$ p=\frac{\pi}{5}(x-1)\in[a,b) $$ for some $a$ and $b$. Find $a$ and $b$ and then find $p$ such that $p\in[a,b)$ and $$ \sin p=\pm\frac{\sqrt{3}}{2}. $$ Once you have the values of $p$ in $[a,b)$, use the relation $$ p=\frac{\pi}{5}(x-1) $$ to the values of $x$.

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For $\sin^2y=\sin^2A\iff\cos^2y=\cos^2A\iff\tan^2y=\tan^2A$

Method$\#1:$ use Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $

$\sin(y\pm A)=0,y=m\pi\mp A$

Method$\#2:$

Use $\cos2x=1-2\sin^2x,$

and $\cos2x=\cos2A\implies2x=2n\pi\pm2A$

lab bhattacharjee
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0

When you get to $sin^2\left({\pi\over5}{(x-1)}\right)={3\over4}$,so using $sin^2(x)+cos^2(x)=1$,$cos\left({\pi\over5}{(x-1)}\right)={1\over2}$,$\left({\pi\over5}{(x-1)}\right)={\pi\over3}+2\pi n$. It is easy from here now.

Math Lover
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