Question

Sin x +√3 coax= √2​

Answer 1

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

 $\frac{1}{2}Sinx + \frac{\sqrt{3} }{2}Cosx = \frac{\sqrt{2} }{2}$

Sin($\frac{pi}{6}$)*Sinx + Cos($\frac{pi}{6}$)*Cosx = $\frac{1}{\sqrt{2} }$

This makes the formula of

CosACosB + SinASinB = Cos(A-B)

 Cos(x-$\frac{pi}{6}$) = $\frac{1}{\sqrt{2} }$

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.