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- Jun 16th 2018, 09:58 PM #1

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- Jun 16th 2018, 10:59 PM #2
## Re: Hai. I have tried the following problem can someone help me.

We are given to solve:

$\displaystyle \left(\sin\left(\frac{x}{4}\right)-2\sin(x)\right)\sin(x)+\left(1+\cos\left(\frac{x}{ 4}\right)-2\cos(x)\right)\cos(x)=0$

I would begin by distributing:

$\displaystyle \sin(x)\sin\left(\frac{x}{4}\right)- 2\sin^2(x)+ \cos(x)+\cos(x)\cos\left(\frac{x}{4}\right)- 2\cos^2(x)=0$

Applying a Pythagorean identity, and product to sum identities, we obtain:

$\displaystyle \frac{1}{2}\left(\cos\left(\frac{3x}{4}\right)- \cos\left(\frac{5x}{4}\right)\right)+ \frac{1}{2}\left(\cos\left(\frac{3x}{4}\right)+ \cos\left(\frac{5x}{4}\right)\right)+ \cos(x)=2$

Combine like terms:

$\displaystyle \cos\left(\frac{3x}{4}\right)+ \cos(x)=2$

At this point, we can see that both arguments must be some integral multiple of $\displaystyle 2\pi$ (why?), which means:

$\displaystyle x=8\pi k$ where $\displaystyle k\in\mathbb{Z}$

- Jun 16th 2018, 11:25 PM #3

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## Re: Hai. I have tried the following problem can someone help me.

Thank you.

sorry for the mistake i did. there is a typing error. the sum is as follows. I tried this in the same way. so i Concluded no solution . Please say am I correct for the problem below.

- Jun 16th 2018, 11:50 PM #4
## Re: Hai. I have tried the following problem can someone help me.

We are given to solve:

$\displaystyle \left(\cos\left(\frac{x}{4}\right)-2\sin(x)\right)\sin(x)+\left(1+\sin\left(\frac{x}{ 4}\right)-2\cos(x)\right)\cos(x)=0$

I would begin by distributing:

$\displaystyle \sin(x)\cos\left(\frac{x}{4}\right)- 2\sin^2(x)+ \cos(x)+\cos(x)\sin\left(\frac{x}{4}\right)- 2\cos^2(x)=0$

Applying a Pythagorean identity, and product to sum identities, we obtain:

$\displaystyle \frac{1}{2}\left(\sin\left(\frac{3x}{4}\right)- \sin\left(\frac{5x}{4}\right)\right)+ \frac{1}{2}\left(\sin\left(\frac{3x}{4}\right)+ \sin\left(\frac{5x}{4}\right)\right)+ \cos(x)=2$

Combine like terms:

$\displaystyle \sin\left(\frac{3x}{4}\right)+ \cos(x)=2$

At this point, we can see that both arguments must be an odd multiple of $\displaystyle \frac{2\pi}{3}$ and at the same time a multiple of $\displaystyle 2\pi$ and so

$\displaystyle x=6\pi(4k+1)$ where $\displaystyle k\in\mathbb{Z}$

- Jun 17th 2018, 12:30 AM #5

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## Re: Hai. I have tried the following problem can someone help me.

$\displaystyle sin(x)cos(x/4)+cos(x)sin(x/4)=sin(5x/4)$

because $\displaystyle sin(A)cos(B)+cos(A)sin(B)=sin(A+B)$

- Jun 17th 2018, 08:26 AM #6

- Jun 17th 2018, 08:11 PM #7
## Re: Hai. I have tried the following problem can someone help me.

To follow up...we see we require:

$\displaystyle \frac{5x}{4}=\frac{\pi}{2}+2\pi k_1=\frac{\pi}{2}(4k_1+1)\implies x=\frac{2\pi}{5}(4k_1+1)$

$\displaystyle x=2\pi k_2$

Thus:

$\displaystyle x=2\pi(4k+1)$