I'll assume the rest of the questions is something like for 0 ≤ x < 2π
Solve for x
[tex]\sin(x+ \frac \pi 4) = \sqrt{2} \cos x[/tex]
Using the sum angle formula for sine,
[tex]\sin x \cos \frac \pi 4 + \cos x \sin \frac \pi 4 = \sqrt{2} \cos x [/tex]
Of course the sine and cosine of π/4 are the same, both √2/2.
[tex]\dfrac{\sqrt{2}}{2}\sin x+\dfrac{\sqrt{2}}{2} \cos x = \sqrt{2} \cos x[/tex]
The square roots of two cancel; let's multiply through by two.
[tex]\sin x + \cos x = 2 \cos x[/tex]
[tex] \sin x = \cos x[/tex]
[tex]\dfrac{\sin x }{\cos x} = 1[/tex]
[tex]\tan x = 1[/tex]
[tex]x = \arctan 1[/tex]
[tex]x = \frac \pi 4 + \pi k, \quad \textrm{integer } k[/tex]
k=0 and k=1 are in the range
Answer: x = π/4 and x = 5π/4
Let's check x=5π/4.
[tex]\sin(\frac{5\pi}{4} + \frac \pi 4) = \sin(\frac{3\pi}{2}) = -1[/tex]
[tex]\sqrt{2} \cos(\frac{5\pi}{4})=\sqrt{2} (-\frac{\sqrt{2}}{2}) = -1 \quad\checkmark[/tex]
