Answer:
The two values of x are 2n*pi + pi/12 and 2n*pi -5pi/12
Explanation:
The given equation is
Sin x +√3 Cosx= √2
Upon dividing the equation by 2 we get
[tex]\frac{1}{2}Sinx + \frac{\sqrt{3} }{2}Cosx = \frac{\sqrt{2} }{2}[/tex]
Sin([tex]\frac{pi}{6}[/tex])*Sinx + Cos([tex]\frac{pi}{6}[/tex])*Cosx = [tex]\frac{1}{\sqrt{2} }[/tex]
This makes the formula of
CosACosB + SinASinB = Cos(A-B)
Cos(x-[tex]\frac{pi}{6}[/tex]) = [tex]\frac{1}{\sqrt{2} }[/tex]
cos(x- pi/6) = cos(pi/4)
upon writing the general equation we get
x-pi/6 = 2n*pi ± pi/4
x = 2n*pi ± pi/4 -pi/6
so we will have two solutions
x = 2n*pi + pi/4 -pi/6
= 2n*pi + pi/12
and
x = 2n*pi - pi/4 -pi/6
= 2n*pi -5pi/12
Therefore the two values of x are 2n*pi + pi/12 and 2n*pi -5pi/12.