 Mary Hammonds

2021-12-31

Find the value of x from $\mathrm{sin}x+\mathrm{cos}x=\sqrt{2}\mathrm{sin}\left(5x\right)$
I used auxiliary argument method and converted into $\mathrm{cos}\left(\frac{\pi }{4}-x\right)=\mathrm{cos}\left(\frac{\pi }{2}-5x\right)$ (by introducing $\sqrt{2}$ as auxiliary argument and then using $\mathrm{cos}\left(\frac{\pi }{4}\right)\mathrm{cos}x+\mathrm{sin}\left(\frac{\pi }{4}\right)\mathrm{sin}x=\mathrm{cos}\left(\frac{\pi }{4}-x\right)$ and using $\mathrm{sin}5x=\mathrm{cos}\left(\frac{\pi }{2}-5x\right)$
But the answer isn't matching after using the formula for $\mathrm{cos}\theta =\mathrm{cos}\alpha$ Travis Hicks

Then
$\frac{\pi }{4}-x=2\pi k±\left(\frac{\pi }{2}-5x\right)$

$=\frac{\pi \left(8n+3\right)}{24}$ vicki331g8

It is much simpler to use congruences.
First note the equation can be written as nick1337

A standard trig formula is
$\mathrm{sin}\left(a+b\right)=\mathrm{sin}a\mathrm{cos}b+\mathrm{sin}b\mathrm{cos}a$
We can use this to say
$\mathrm{sin}x+\mathrm{cos}x=\sqrt{2}\left(\mathrm{sin}x×\frac{1}{\sqrt{2}}+\mathrm{cos}x×\frac{1}{\sqrt{2}}\right)=\sqrt{2}\left(\mathrm{sin}x\mathrm{cos}\left(\frac{\pi }{4}\right)+\mathrm{cos}x\mathrm{sin}\left(\frac{\pi }{4}\right)\right)=\sqrt{2}\mathrm{sin}\left(x+\frac{\pi }{4}\right)$
So
$\sqrt{2}\mathrm{sin}\left(x+\frac{\pi }{4}\right)=\sqrt{2}\mathrm{sin}\left(5x\right)$
Therefore, $x+\frac{\pi }{4}=5x$ (modulo $2\pi$). And you should be able to take it from there.
Peterwhy's comment reminds me, you also need to take into account that

Do you have a similar question?