How do I solve for the value of x in degrees if √2sinx-1=0?

1 Answer
Feb 21, 2018

x=135°+360°*n or x=360°*n+45°

Explanation:

First, let's solve it in radians

sqrt2sinx-1=0
therefore sinx=1/sqrt2=sqrt2/2
If you look at Special Ratio, sin45° is sqrt2/2
proof)
enter image source here
SOH CAH TOA, sin45°=(Opposite)/(Hypotenuse)
therefore 1/sqrt2=sqrt2/2

But this was the only case when 0°<=x<=90°(0<=x<=pi/2)

Now we need to look on real number range

Since sin has period of 2pi, (f(?)=f(?+2pi))

thereforex=360°*n+45°(2npi+pi/4)

But we know that sin function is positive @ 2nd quadrant.(-, +) Which means,
enter image source here
We need to find out the sinx=sqrt2/2 for 2nd quadrant (90°<=x<=180°(pi/2<=x<=pi))
x=(3pi)/4
But sinx has period of 2pi,

therefore x=(3pi)/4+2npi

If you write them in degrees,

x=(3*180°)/4+360°*n=135°+360°*n

therefore x=135°+360°*n